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.
def f(al,ii):
    st=False
    cr=0
    for i in al:
        i.sort()
        if not st and (ii<i[0] or ii>i[1]):
            continue
        else:
            st=True
            if cr==0:
                cr=i
            #print(cr,i)
            tpp=[cr,i]
            tpp.sort()
            if tpp[-1][0]<=tpp[0][1]:
                tp=merge(tpp)
                if len(tp)==1:
                    tp[0].sort()
                    cr=tp[0]
    return cr
    
    
def merge(intervals):
 
    intervals.sort(key=lambda x: x[0])
 
    merged = []
    for interval in intervals:
        # if the list of merged intervals is empty or if the current
        # interval does not overlap with the previous, simply append it.
        if not merged or merged[-1][1] < interval[0]:
            merged.append(interval)
        else:
        # otherwise, there is overlap, so we merge the current and previous
        # intervals.
            merged[-1][1] = max(merged[-1][1], interval[1])
 
    return merged
    
    
for tc in range(int(input())):
    al=list(map(int,input().split()))
    x=al[1]
    m=al[2]
    al=[]
    for j in range(m):
        al.append(list(map(int,input().split())))
    tt=(f(al,x))
    if tt==0:
        print(1)
    else:
        print(tt[1]-tt[0]+1)