CF2255E1 What Will Remain at the End? (Easy Version)

This is the easy version of the problem. The only difference between the two versions is the set of allowed values for the initial array and for $ x $ in operations of type $ 1 $ . In this version, all these values belong to $ \{-1,0,1\} $ . You can make hacks only if both versions of the problem are solved. Before her final sortie, Chtholly asks Willem three questions. The second is this: what will remain if the sky truly reaches its end? Willem cannot answer her directly. Instead, he opens a chronicle containing $ n $ records, numbered from $ 1 $ to $ n $ . Each record holds an integer: a positive value represents hope, while a negative value represents despair. The initial contents of the chronicle form an array $ a_1,a_2,\ldots,a_n $ , called version $ 0 $ . Chtholly then performs $ q $ operations. For each $ 1\le i\le q $ , the $ i $ -th operation creates a new version $ i $ from version $ i-1 $ . Each operation has one of the following four types: - $ \texttt{1 l r x} $ : set $ a_k\gets x $ for every $ l\le k\le r $ . - $ \texttt{2 l r} $ : set $ a_k\gets -a_k $ for every $ l\le k\le r $ . - $ \texttt{3 l r} $ : set $ a_k\gets\max(a_k,0) $ for every $ l\le k\le r $ . - $ \texttt{4 p} $ : consider the value at position $ p $ in every previous version $ 0,1,\ldots, i-1 $ . Let these values be $ b_0,b_1,\ldots,b_{i-1} $ . Find the maximum sum over all non-empty subarrays $ ^{\text{∗}} $ of this sequence. If the operation is of type $ 1 $ , $ 2 $ , or $ 3 $ , the specified modification is applied to version $ i-1 $ to obtain version $ i $ . An operation of type $ 4 $ does not modify the array, so version $ i $ is identical to version $ i-1 $ . The operations are encoded and must be processed in order. Their decoding depends on $ \mathrm{lastans} $ , which is updated after every operation of type $ 4 $ . Help Willem answer every operation of type $ 4 $ . $ ^{\text{∗}} $ An array $ c $ is a subarray of an array $ b $ if $ c $ can be obtained from $ b $ by the deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1 \le t \le 10^4 $ ). The description of the test cases follows. The first line of each test case contains two integers $ n $ and $ q $ ( $ 1\le n,q\le5\cdot10^5 $ ) — the length of the array and the number of operations. The second line of each test case contains $ n $ integers $ a_1,a_2,\ldots,a_n $ ( $ a_i\in\{-1,0,1\} $ ) — the array in version $ 0 $ . Each of the next $ q $ lines describes one operation in one of the following encoded formats: The first integer on the line is the operation type. - $ \texttt{1 u v x} $ ( $ 0\le u,v \lt 2^{64} $ , $ x\in\{-1,0,1\} $ ); - $ \texttt{2 u v} $ ( $ 0\le u,v \lt 2^{64} $ ); - $ \texttt{3 u v} $ ( $ 0\le u,v \lt 2^{64} $ ); - $ \texttt{4 u} $ ( $ 0\le u \lt 2^{64} $ ). The operations are encoded and must be processed in order. Their decoding depends on a value $ \mathrm{lastans} $ . Initially, $ \mathrm{lastans}=0 $ . After answering an operation of type $ 4 $ , set $ \mathrm{lastans} $ to the least non-negative residue of its answer modulo $ 2^{64} $ . Operations of all other types leave $ \mathrm{lastans} $ unchanged. For every encoded coordinate $ y $ , define $ d(y)=\left(\left(y\oplus\mathrm{lastans}\right)\bmod n\right)+1 $ . Here, $ \oplus $ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). - For an operation $ \texttt{1 u v x} $ , let $ l=\min(d(u),d(v)) $ and $ r=\max(d(u),d(v)) $ . The value $ x $ is not encoded. - For an operation $ \texttt{2 u v} $ or $ \texttt{3 u v} $ , let $ l=\min(d(u),d(v)) $ and $ r=\max(d(u),d(v)) $ . - For an operation $ \texttt{4 u} $ , let $ p=d(u) $ . The operation type is not encoded. Do not forget to update $ \mathrm{lastans} $ after answering each operation of type $ 4 $ . It is guaranteed that the sum of $ n $ over all test cases does not exceed $ 5\cdot10^5 $ . It is guaranteed that the sum of $ q $ over all test cases does not exceed $ 5\cdot10^5 $ .

For every operation of type $ 4 $ , output a single integer — the maximum sum of a non-empty subarray of the sequence of the values at position $ p $ over all versions preceding this operation.

In the first test case, for the first operation, $ \mathrm{lastans}=0 $ , so the encoded coordinate $ 1 $ is decoded into position $ 2 $ . Only version $ 0 $ is considered. The value at position $ 2 $ is $ -1 $ , hence the answer is $ -1 $ . Now $ \mathrm{lastans}=2^{64}-1=18\,446\,744\,073\,709\,551\,615 $ . Therefore the encoded interval $ [18\,446\,744\,073\,709\,551\,615,18\,446\,744\,073\,709\,551\,613] $ is decoded into $ [1,3] $ , and the encoded coordinate $ 18\,446\,744\,073\,709\,551\,614 $ is decoded into position $ 2 $ . Before the second query, the values at position $ 2 $ in versions $ 0 $ , $ 1 $ , and $ 2 $ are $ -1 $ , $ -1 $ , and $ 1 $ , respectively. Their maximum non-empty subarray sum is $ 1 $ . Before the third query, the values at position $ 2 $ in versions $ 0,1,\ldots,5 $ form sequence $ [-1,-1,1,1,1,-1] $ . The three consecutive values equal to $ 1 $ form a subarray with sum $ 3 $ . Notice that the first, second, and third queries create versions $ 1 $ , $ 3 $ , and $ 6 $ , respectively, even though they do not change the array. In the second test case, the first query asks about position $ 1 $ , so its answer is $ 1 $ . Then $ \mathrm{lastans}=1 $ , and the encoded operation $ \texttt{1 0 3 -1} $ is decoded into $ \texttt{1 2 3 -1} $ . Before the second query, the values at position $ 3 $ in versions $ 0 $ , $ 1 $ , and $ 2 $ are $ 1 $ , $ 1 $ , and $ -1 $ , respectively, so the answer is $ 2 $ . Afterwards, $ \mathrm{lastans}=2 $ . The encoded operations $ \texttt{2 2 3} $ and $ \texttt{3 3 0} $ are decoded into $ \texttt{2 1 2} $ and $ \texttt{3 2 3} $ , respectively. Before the last query, the values at position $ 2 $ in versions $ 0,1,\ldots,5 $ form sequence $ [-1,-1,-1,-1,1,1] $ , whose maximum non-empty subarray sum is $ 2 $ .