You are given an array consisting of n integers a1, a2, …, an. Initially ax=1, all other elements are equal to 0.

You have to perform m operations. During the i-th operation, you choose two indices c and d such that li≤c,d≤ri, and swap ac and ad.

Calculate the number of indices k such that it is possible to choose the operations so that ak=1 in the end.

Input
The first line contains a single integer t (1≤t≤100) — the number of test cases. Then the description of t testcases follow.

The first line of each test case contains three integers n, x and m (1≤n≤109; 1≤m≤100; 1≤x≤n).

Each of next m lines contains the descriptions of the operations; the i-th line contains two integers li and ri (1≤li≤ri≤n).

Output
For each test case print one integer — the number of indices k such that it is possible to choose the operations so that ak=1 in the end.

Example
Input
3
6 4 3
1 6
2 3
5 5
4 1 2
2 4
1 2
3 3 2
2 3
1 2
Output
6
2
3
Note
In the first test case, it is possible to achieve ak=1 for every k. To do so, you may use the following operations:

swap ak and a4;
swap a2 and a2;
swap a5 and a5.
In the second test case, only k=1 and k=2 are possible answers. To achieve a1=1, you have to swap a1 and a1 during the second operation. To achieve a2=1, you have to swap a1 and a2 during the second operation.
思路：只要區間中包含1，那么這一個區間中所有的位置都有可能為1.設置左右兩個指針，不斷的更新有可能為1的區間長度就可以了。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e2+10; struct node{int x,y; }p[maxx]; int n,x,m;int main() {int t;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&x,&m);int l=x,r=x;for(int i=1;i<=m;i++){scanf("%d%d",&p[i].x,&p[i].y);if(!(p[i].x>r||p[i].y<l)) l=min(l,p[i].x),r=max(r,p[i].y);}cout<<r-l+1<<endl;}return 0; }

努力加油a啊，(^o)/~

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