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.
"""
 Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
 
def ceil(x):
    if x != int(x):
        x = int(x) + 1;
    return x;

def factorial(x, m):
	val = 1
	while x>0:
		val = (val * x) % m
		x -= 1
	return val
    
# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;
 
## gcd function
def gcd(a,b):
    if b == 0:
        return a;
    return gcd(b, a % b);
 
## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k; 
    res = 1;
    for i in range(k): 
        res = res * (n - i);
        res = res / (i + 1); 
    return int(res);
 
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
    if hi == None:
        hi = len(a);
    while lo < hi:
        mid = (lo+hi)//2;
        if a[mid] < x:
            lo = mid+1;
        else:
            hi = mid;
    return lo;
 
## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes
 
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 
 
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b;
 
# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];
 
# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;
 
 
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x
 
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime
 
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array
 
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret;
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)
 
## taking integer array input
def int_array():
    return list(map(int, input().strip().split()));
 
def float_array():
    return list(map(float, input().strip().split()));
 
## taking string array input
def str_array():
    return input().strip().split();
 
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
 
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
def solve():
	n, x, m = map(int, input().split())
	pos = []
	l1, r1 = x, x
	for i in range(m):
		l, r = map(int, input().split())
		# print(l, r, x)
		if l > r1 or r < l1:
			continue
		else:
			l1, r1 = min(l, l1), max(r, r1)
	print(r1 - l1 + 1)

			

if __name__ == '__main__':
	for _ in range(int(input())):
		solve()
	# fin_time = datetime.now()
# 	print("Execution time (for loop): ", (fin_time-init_time))

