You are given two integers a and b, and q queries. The i-th query consists of two numbers l_i and r_i, and the answer to it is the number of integers x such that l_i ≤ x ≤ r_i, and ((x mod a) mod b) ≠ ((x mod b) mod a). Calculate the answer for each query.

Recall that y mod z is the remainder of the division of y by z. For example, 5 mod 3 = 2, 7 mod 8 = 7, 9 mod 4 = 1, 9 mod 9 = 0.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then the test cases follow.

The first line of each test case contains three integers a, b and q (1 ≤ a, b ≤ 200; 1 ≤ q ≤ 500).

Then q lines follow, each containing two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^{18}) for the corresponding query.

Output

For each test case, print q integers — the answers to the queries of this test case in the order they appear.

Example

Input


2
4 6 5
1 1
1 3
1 5
1 7
1 9
7 10 2
7 8
100 200


Output


0 0 0 2 4 
0 91 
# -*- coding: utf-8 -*-
import sys, string
from collections import deque, defaultdict
from math import sqrt, factorial, gcd, ceil, atan, pi, log2
# def input(): return sys.stdin.readline()[:-1] # warning not \n
# def input(): return sys.stdin.buffer.readline().strip() # warning bytes
def input(): return sys.stdin.buffer.readline().decode('utf-8')
# string.ascii_lowercase
from bisect import bisect_left, bisect_right
from operator import add
MOD = int(1e9)+7
INF = float('inf')

# sys.setrecursionlimit(int(1e6))




def solve():
    # n = int(input())
    a, b, q = [int(x) for x in input().split()]
    n = a * b + 1
    m = [0] * n
    for x in range(1, n):
        if (x % a) % b != (x % b) % a:
            m[x] = 1
        m[x] += m[x - 1]

    cur = [0] * q
    for i in range(q):
        l, r = [int(x) for x in input().split()]
        l -= 1
        ans = r // (a * b) * m[a * b] + m[r % (a * b)]
        ans -= l // (a * b) * m[a * b] + m[l % (a * b)]
        cur[i] = ans
    print(*cur)
    
    

T = 1
T = int(input())
for case in range(1,T+1):
    # try:
        ans = solve()
    # except:
    #     import traceback
    #     print(traceback.format_exc())

"""

5

1 1
1 1 1
2 2 2 2

0 0 1 1 2
2 0 1 0 0

1 

2

"""

