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 
from sys import  stdin, stdout
from math import *
from heapq import *
from collections import *
        
def main():
    ntest=int(stdin.readline())
    for testcase in range(ntest):
        a,b,q=[int(x) for x in stdin.readline().split()]
        f={-1:0}
        for i in range((a*b)+2):
            f[i]=f[i-1]+(0 if (((i%a)%b)==((i%b)%a)) else  1)
        F=lambda n: ((f[a*b]*(n//(a*b))) + f[n%(a*b)])
        for case in range(q):
            l,r=[int(x) for x in stdin.readline().split()]
            stdout.write("%d "%(F(r)-F(l-1)))
        stdout.write("\n")
    return 0

if __name__ == "__main__":
    main()