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 math as mt 
import sys,string
input=sys.stdin.readline
#print=sys.stdout.write
#import random
from collections import defaultdict
from heapq import heappush,heapify,heappop
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())

def check(a,b,c,d):
    if(a==c or a==d) or (b==c or b==d):
        return True
    elif(a<=c and c<=b) or (a<=d and d<=b):
        return True
    elif(c<=a and a<=d) or (c<=b and b<=d):
        return True
    return False
t=I()

right=0
left=10**10

    

for _ in range(t):
    n,x,m=M()
    f=0
    p=10**10
    q=-1
    d=[]
    for i in range(m):
        a,b=M()
        d.append((a,b))
        if((x>=a and x<=b) or ((x==a) or (x==b))) and f==0:
            key=i
            p=a
            q=b
            f=1
    #print(key,p,q)
    if(p==10**10 and q==-1):
        print(1)
    else:
        for j in range(key,m):

            if(check(p,q,d[j][0],d[j][1])):
                p=min(p,d[j][0])
                q=max(q,d[j][1])
        print(q-p+1)
            
