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.
t=int(input())
i=0
while(i<t):
    summer=0
    n,x,m=list(map(int,input().split()))
    j=0
    ls=[[x,x]]
    while(j<m):
        notFound=True
        l,r=list(map(int,input().split()))
        k=0
        while(k <len(ls) and notFound):
            if((ls[k][0]>=l and ls[k][0]<=r) or (ls[k][1]>=l and ls[k][1]<=r)):
                notFound=False
            k+=1

        if(not notFound):

            ls.append([l,r])



        j+=1
    # print(f'the conjoint list={ls}')    


    ls.sort(key=lambda s:s[0])
    # print(f'the conjoint list after sorting={ls}')    

    j=0
    while(j<len(ls)):
        k=j+1
        # print(f'j={j}, k={k}, len={len(ls)}, ls={ls}')
        if(k==len(ls)):
            j+=1
        else:
            # print(f'ls[j]={ls[j]},     ls[k]={ls[k]}')

            if(ls[k][0]<=ls[j][1]):
                ls[j][1]=max(ls[j][1],ls[k][1])
                ls[k:]=ls[k+1:]
            else:
                j+=1
        
        

    # print(f'the conjoint list after merging={ls}')    
    summer=0
    for l in ls:
        summer+=(l[1]-l[0]+1)
    print(summer)


    
    i+=1