CF1343A Candies

Recently Vova found $ n $ candy wrappers. He remembers that he bought $ x $ candies during the first day, $ 2x $ candies during the second day, $ 4x $ candies during the third day, $ \dots $ , $ 2^{k-1} x $ candies during the $ k $ -th day. But there is an issue: Vova remembers neither $ x $ nor $ k $ but he is sure that $ x $ and $ k $ are positive integers and $ k > 1 $ . Vova will be satisfied if you tell him any positive integer $ x $ so there is an integer $ k>1 $ that $ x + 2x + 4x + \dots + 2^{k-1} x = n $ . It is guaranteed that at least one solution exists. Note that $ k > 1 $ . You have to answer $ t $ independent test cases.

The first line of the input contains one integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. Then $ t $ test cases follow. The only line of the test case contains one integer $ n $ ( $ 3 \le n \le 10^9 $ ) — the number of candy wrappers Vova found. It is guaranteed that there is some positive integer $ x $ and integer $ k>1 $ that $ x + 2x + 4x + \dots + 2^{k-1} x = n $ .

Print one integer — any positive integer value of $x$ so there is an integer $k>1$ that $x+2x+4x+⋯+2^{k-1}x=n$.

In the first test case of the example, one of the possible answers is $ x=1, k=2 $ . Then $ 1 \cdot 1 + 2 \cdot 1 $ equals $ n=3 $ . In the second test case of the example, one of the possible answers is $ x=2, k=2 $ . Then $ 1 \cdot 2 + 2 \cdot 2 $ equals $ n=6 $ . In the third test case of the example, one of the possible answers is $ x=1, k=3 $ . Then $ 1 \cdot 1 + 2 \cdot 1 + 4 \cdot 1 $ equals $ n=7 $ . In the fourth test case of the example, one of the possible answers is $ x=7, k=2 $ . Then $ 1 \cdot 7 + 2 \cdot 7 $ equals $ n=21 $ . In the fifth test case of the example, one of the possible answers is $ x=4, k=3 $ . Then $ 1 \cdot 4 + 2 \cdot 4 + 4 \cdot 4 $ equals $ n=28 $ .