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 
# python3

##=====================================================
from sys import stdin, stdout
from collections import defaultdict, Counter
from functools import lru_cache
from math import gcd, floor, ceil
 
def ilist():
    return [int(x) for x in stdin.readline().strip().split(" ")]
def iint():
    return int(stdin.readline().strip())
def istr():
    return stdin.readline().strip()
##=====================================================


if __name__ == '__main__':
    t = iint()
    for _ in range(t):
        a, b, q = ilist()
        res = []
        x = 1
        lcm = (a * b) // gcd(a, b)
        p = [0] * (lcm)
        for x in range(1, lcm):
            if ((x % a) % b) != ((x % b) % a):
                p[x] = 1
            p[x] += p[x-1]
        for _ in range(q):
            l, r = ilist()
            s = 0
            l -= 1
            s += (r // lcm) * p[-1] + p[r % lcm]
            s -= (l // lcm) * p[-1] + p[l % lcm]
            res.append(str(s))
        print(" ".join(res))
