CF Polycarpus' Dice (数学)

Polycarpus' Dice

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

2 8
4 4

output

3 3 

input

1 3
5

output

4 

input

2 3
2 3

output

0 1 



当前骰子最小值 + 剩余筛子最大值 >= A
当前骰子最大值 + 剩余筛子最小值 <= A

#include <iostream>

#include <cmath>

#include <cstring>

#include <cstdio>

#include <string>

#include <algorithm>

#include <cctype>

#include <queue>

#include <map>

using    namespace    std;



int    main(void)

{

    int        n;

    int        s[200005];

    long    long    A;

    long    long    sum = 0;

    long    long    ans = 0;



    cin >> n >> A;

    for(int i = 0;i < n;i ++)

    {

        cin >> s[i];

        sum += s[i];

    }

    for(int i = 0;i < n;i ++)

    {

        long    long    m_min = A - sum + s[i];

        long    long    m_max = A - (n - 1);

        if(m_max > s[i])

            m_max = s[i];

        ans = s[i] - m_max + (m_min - 1 > 0 ? m_min - 1: 0);

        cout << ans;

        if(i != n - 1)

            cout << ' ';

    }

    cout << endl;



    return    0;

}