You are given an array consisting of n integers a_1, a_2, ..., a_n. Initially a_x = 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 l_i ≤ c, d ≤ r_i, and swap a_c and a_d.

Calculate the number of indices k such that it is possible to choose the operations so that a_k = 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 ≤ 10^9; 1 ≤ m ≤ 100; 1 ≤ x ≤ n).

Each of next m lines contains the descriptions of the operations; the i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 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 a_k = 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 a_k = 1 for every k. To do so, you may use the following operations:

  1. swap a_k and a_4; 
  2. swap a_2 and a_2; 
  3. swap a_5 and a_5. 



In the second test case, only k = 1 and k = 2 are possible answers. To achieve a_1 = 1, you have to swap a_1 and a_1 during the second operation. To achieve a_2 = 1, you have to swap a_1 and a_2 during the second operation.
import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t -- > 0) {
            int n = sc.nextInt(); int x = sc.nextInt(); int m = sc.nextInt();
            Pairs[] pair = new Pairs[m];
            for(int i=0;i<m;i++){
                int start = sc.nextInt(); int end = sc.nextInt();
                pair[i] = new Pairs(start, end);
            }
            int ans = 1;
            for (int i=0;i<m;i++){
                if(pair[i].start <= x && pair[i].end >=x){
                    ans = Math.max(ans, pair[i].end-pair[i].start+1);
                    for(int j=i+1;j< m;j++){
                        Pairs[] tmp = new Pairs[]{pair[i], pair[j]};
                        Arrays.sort(tmp);
                        int start1 = tmp[0].start; int end1 = tmp[0].end;
                        int start2 = tmp[1].start; int end2 = tmp[1].end;
                        int end = Math.max(end1,end2);
                        if(start2 <= end1) {
                            pair[j] = new Pairs(Math.min(start1, pair[i].start), Math.max(pair[i].end, end));
                            ans = Math.max(ans, pair[j].end-pair[j].start+1);
                        }
                    }
                }
            }

            System.out.println(ans);
        }
    }

    private static class Pairs implements Comparable<Pairs>{
        int start; int end;

        public Pairs(int start, int end) {
            this.start = start;
            this.end = end;
        }

        @Override
        public int compareTo(Pairs pairs) {
            if(this.start==pairs.start){
                return Integer.compare(this.end,pairs.end);
            }else{
                return Integer.compare(this.start, pairs.start);
            }
        }
    }
}
