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 
# cook your dish here
from sys import stdin, stdout
import math
from itertools import permutations, combinations
from collections import defaultdict
from bisect import bisect_left 
from bisect import bisect_right
 
def L():
    return list(map(int, stdin.readline().split()))
 
def In():
    return map(int, stdin.readline().split())
 
def I():
    return int(stdin.readline())
 
P = 1000000007
    
def main():
    for t in range(I()):
        a, b, q = In()
        pd = a*b
        arr = [0 for i in range(pd)]
        for i in range(1, pd):
            arr[i] = arr[i-1] + (((i)%a)%b != ((i)%b)%a)
        ans = []
        for i in range(q):
            l, r = In()
            val = (r//pd)*arr[-1] + arr[r%pd]
            val -= ((l-1)//pd)*arr[-1] + arr[(l-1)%pd]
            ans.append(val)
        print(*ans)
        
    
if __name__ == '__main__':
    main()