SDKD 2016 Summer Single Contest #12 .G

Polycarp has n dice d1, d2, …, dn. The i-th dice shows numbers from 1 to di. 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 d1, d2, …, dn. 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.

Input

The first line contains two integers n, A (1 ≤ n ≤ 2·105, n ≤ A ≤ s) — the number of dice and the sum of shown values where s = d1 + d2 + … + dn.

The second line contains n integers d1, d2, …, dn (1 ≤ di ≤ 106), where di is the maximum value that the i-th dice can show.

Output

Print n integers b1, b2, …, bn, where bi is the number of values for which it is guaranteed that the i-th dice couldn’t show them.

Sample Input

Input

2 8

4 4

Output

3 3

Input

1 3

5

Output

4

Input

2 3

2 3

Output

0 1

Hint

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.

题目：

这就是一数学题，还要注意 longlong

这题就是 要求一个范围 即 每个dn 可以 取得范围 ，其他的 就是不能取得。

现在求范围：

除了这个数 其他的数 取 最小 1 那么这个数就要取得最大 ，

除了这个数 其他的数 取 最大 sum - dn 那么这个数就取最小

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 200000 + 100;
long long n, A;
long long sum = 0;
long long d[maxn];
int main()
{
cin >> n >> A;
for(int i = 0; i < n; i++)
{
cin >> d[i];
sum += d[i];
}
for(int i = 0; i < n; i++)
{
int mmax = min(A - (n-1), d[i]);
int mmin = max(1LL, A - (sum - d[i]));
if(i != 0)
printf(" ");
printf("%d", mmin-1 + d[i] - mmax);
}
return 0;
}