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.
#include <CodeforcesSolutions.h>
#include <ONLINE_JUDGE <solution.cf(contestID = "1362",problemID = "B",method = "GET")>.h>
"""
    Author : thekushalghosh
    Team   : CodeDiggers
 
    I prefer Python language over the C++ language :p :D
        
    Visit my website : thekushalghosh.github.io
"""
import sys,math,cmath,time,random
start_time = time.time()
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################

def solve():
    n,x,m = invr()
    c = 0
    a = []
    q = -1
    w = -1
    qw = 0
    for i in range(m):
        l,r = invr()
        if l <= x <= r and qw == 0:
            q = l
            w = r
            qw = 1
        if qw == 1 and not(l > w or r < q):
            q = min(q,l)
            w = max(w,r)
    if qw == 0:
        c = 1
    else:
        c = w - q + 1
    print(c)
                
################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
 
def main():
    global tt
    if not ONLINE_JUDGE:
        sys.stdin = open("input.txt","r")
        sys.stdout = open("output.txt","w")
    t = 1
    t = inp()
    for tt in range(t):
        solve()
    if not ONLINE_JUDGE:
        print("Time Elapsed :",time.time() - start_time,"seconds")
        sys.stdout.close()
    
#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    return(input().strip())
def invr():
    return(map(int,input().split()))
 
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def counter(a):
    q = [0] * max(a)
    for i in range(len(a)):
        q[a[i] - 1] = q[a[i] - 1] + 1
    return(q)
 
def string_counter(a):
    q = [0] * 26
    for i in range(len(a)):
        q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
    return(q)

def power_two(n):
    if n == 1:
        return(True)
    sttr = str(n)
    len_str = len(sttr)
    sttr = list(sttr)
    num = 0
    if (len_str == 1 and sttr[len_str - 1] == '1'): 
        return(False)
    while (len_str != 1 or sttr[len_str - 1] != '1'):
        if ((ord(sttr[len_str - 1]) - ord('0')) % 2 == 1):
            return(False)
        j = 0; 
        for i in range(len_str):  
            num = num * 10 + (ord(sttr[i]) - ord('0')); 
            if (num < 2): 
                if (i != 0):  
                    sttr[j] = '0'
                    j += 1
                continue
  
            sttr[j] = chr((num // 2) + ord('0'))
            j += 1; 
            num = (num) - (num // 2) * 2
        len_str = j
    return(True)

def log_two(n): 
    return(math.log(n) / math.log(2))

#-----------------------------------------------------------------------#
        
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
    input = sys.stdin.readline
   
    
main()