You are the gym teacher in the school.

There are nn students in the row. And there are two rivalling students among them. The first one is in position aa, the second in position bb. Positions are numbered from 11 to nn from left to right.

Since they are rivals, you want to maximize the distance between them. If students are in positions pp and ss respectively, then distance between them is |p−s||p−s|.

You can do the following operation at most xx times: choose two adjacent (neighbouring) students and swap them.

Calculate the maximum distance between two rivalling students after at most xx swaps.

The first line contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.

The only line of each test case contains four integers nn, xx, aa and bb (2≤n≤1002≤n≤100, 0≤x≤1000≤x≤100, 1≤a,b≤n1≤a,b≤n, a≠ba≠b) — the number of students in the row, the number of swaps which you can do, and positions of first and second rivaling students respectively.

For each test case print one integer — the maximum distance between two rivaling students which you can obtain.

In the first test case you can swap students in positions 33 and 44. And then the distance between the rivals is equal to |4−2|=2|4−2|=2.

In the second test case you don't have to swap students.

In the third test case you can't swap students.

思路：情况1：两个人的距离加上可移动的次数都小于等于最远距离的话，直接输出。

情况2：两个人的距离加上可移动距离大于最远距离，则说明可移动次数x足够用了。再来判断，要使两个人达到最远距离，与最远距离还差多少，如果可移动次数x大于差值，则两人的距离可达最大，否则两个人的最远距离为原先的距离加上可移动距离x，即为答案

AC代码

#include<iostream>

#include<cmath>

using namespace std;

int main()

{

    int t;

    cin >> t;

    while(t--)

    {

        int juli,n,x,a,b;

        cin >> n >> x >> a >> b;

        juli = abs(a - b);

        if(juli + x <= n - )//移动后的距离 小于等于 最远距离

        {

            cout << x + juli << endl;//直接输出

            continue;

        }

        if(n - juli <= x)//n-juli为 离最远距离差多少

        {

            cout << n -  << endl;

            continue;

        }

        else

        {

            cout << juli + x << endl;

            continue;

        }

    }

    return ;

}

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