Polycarpus' Dice

题意翻译

有 $n$ 个骰子，第 $i$ 个骰子朝上的面的可能值在 $1$ 到 $d_i$ 之间，扔了一遍这些骰子之后发现朝上的面的值之和为 $A$。现在需要你对于每一个骰子计算出其抛出多少种值一定凑不成 $A$。$1\le n\le 2\times10^5$,$n\le A\le \sum d_i$,$1\le d_i\le 10^6$

题目描述

Polycarp has $ n $ dice $ d_{1},d_{2},...,d_{n} $ . The $ i $ -th dice shows numbers from $ 1 $ to $ d_{i} $ . Polycarp rolled all the dice and the sum of numbers they showed is $ A $ . Agrippina didn't see which dice showed what number, she knows only the sum $ A $ and the values $ d_{1},d_{2},...,d_{n} $ . However, she finds it enough to make a series of statements of the following type: dice $ i $ couldn't show number $ r $ . For example, if Polycarp had two six-faced dice and the total sum is $ A=11 $ , then Agrippina can state that each of the two dice couldn't show a value less than five (otherwise, the remaining dice must have a value of at least seven, which is impossible). For each dice find the number of values for which it can be guaranteed that the dice couldn't show these values if the sum of the shown values is $ A $ .

输入输出格式

输入格式

The first line contains two integers $ n,A $ ( $ 1<=n<=2·10^{5},n<=A<=s $ ) — the number of dice and the sum of shown values where $ s=d_{1}+d_{2}+...+d_{n} $ . The second line contains $ n $ integers $ d_{1},d_{2},...,d_{n} $ ( $ 1<=d_{i}<=10^{6} $ ), where $ d_{i} $ is the maximum value that the $ i $ -th dice can show.

输出格式

Print $ n $ integers $ b_{1},b_{2},...,b_{n} $ , where $ b_{i} $ is the number of values for which it is guaranteed that the $ i $ -th dice couldn't show them.

输入输出样例

输入样例 #1

2 8
4 4

输出样例 #1

3 3 

输入样例 #2

1 3
5

输出样例 #2

4 

输入样例 #3

2 3
2 3

输出样例 #3

0 1 

说明

In the first sample from the statement $ A $ equal to 8 could be obtained in the only case when both the first and the second dice show 4. Correspondingly, both dice couldn't show values 1, 2 or 3. In the second sample from the statement $ A $ equal to 3 could be obtained when the single dice shows 3. Correspondingly, it couldn't show 1, 2, 4 or 5. In the third sample from the statement $ A $ equal to 3 could be obtained when one dice shows 1 and the other dice shows 2. That's why the first dice doesn't have any values it couldn't show and the second dice couldn't show 3.