Antenna Coverage

街道上有$n$个天线。第$i$个天线的位置为$x_i$，以及一个范围值$s_i$；第$i$个天线的覆盖范围是$[x_i-s_i,x_i+s_i]$。 每次操作，你可以花费$1$代价，使得第$i$个天线的$s_i$增加一。每个天线都可以进行多次操作。现在请问你最少需要花费多少代价，使$[1,m]$编号内的每一个位置都被至少一个天线覆盖。

The mayor of the Central Town wants to modernize Central Street, represented in this problem by the $ (Ox) $ axis. On this street, there are $ n $ antennas, numbered from $ 1 $ to $ n $ . The $ i $ -th antenna lies on the position $ x_i $ and has an initial scope of $ s_i $ : it covers all integer positions inside the interval $ [x_i - s_i; x_i + s_i] $ . It is possible to increment the scope of any antenna by $ 1 $ , this operation costs $ 1 $ coin. We can do this operation as much as we want (multiple times on the same antenna if we want). To modernize the street, we need to make all integer positions from $ 1 $ to $ m $ inclusive covered by at least one antenna. Note that it is authorized to cover positions outside $ [1; m] $ , even if it's not required. What is the minimum amount of coins needed to achieve this modernization?

The first line contains two integers $ n $ and $ m $ ( $ 1 \le n \le 80 $ and $ n \le m \le 100\ 000 $ ). The $ i $ -th of the next $ n $ lines contains two integers $ x_i $ and $ s_i $ ( $ 1 \le x_i \le m $ and $ 0 \le s_i \le m $ ). On each position, there is at most one antenna (values $ x_i $ are pairwise distinct).

You have to output a single integer: the minimum amount of coins required to make all integer positions from $ 1 $ to $ m $ inclusive covered by at least one antenna.

3 595
43 2
300 4
554 10

281

1 1
1 1

0

2 50
20 0
3 1

30

5 240
13 0
50 25
60 5
155 70
165 70

26

In the first example, here is a possible strategy: - Increase the scope of the first antenna by $ 40 $ , so that it becomes $ 2 + 40 = 42 $ . This antenna will cover interval $ [43 - 42; 43 + 42] $ which is $ [1; 85] $ - Increase the scope of the second antenna by $ 210 $ , so that it becomes $ 4 + 210 = 214 $ . This antenna will cover interval $ [300 - 214; 300 + 214] $ , which is $ [86; 514] $ - Increase the scope of the third antenna by $ 31 $ , so that it becomes $ 10 + 31 = 41 $ . This antenna will cover interval $ [554 - 41; 554 + 41] $ , which is $ [513; 595] $ Total cost is $ 40 + 210 + 31 = 281 $ . We can prove that it's the minimum cost required to make all positions from $ 1 $ to $ 595 $ covered by at least one antenna. Note that positions $ 513 $ and $ 514 $ are in this solution covered by two different antennas, but it's not important. — In the second example, the first antenna already covers an interval $ [0; 2] $ so we have nothing to do. Note that the only position that we needed to cover was position $ 1 $ ; positions $ 0 $ and $ 2 $ are covered, but it's not important.