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 math import floor, ceil

T = int(input())
for _ in range(T):
    a, b, q = [int(e) for e in input().split(" ")]
    ab = a * b
    sat = []
    num = 0
    for i in range(ab):
        if ((i % a) % b) != ((i % b) % a):
            sat.append(True)
            num += 1
        else:
            sat.append(False)
    
    pat = [0] * ab
    for i in range(1, ab):
        pat[i] = pat[i-1] + (1 if sat[i] else 0)

    
    for _ in range(q):
        l, r = [int(e) for e in input().split(" ")]
        if r - l <= ab:
            tum = 0
            for i in range(l, r+1):
                if ((i % a) % b) != ((i % b) % a):    
                    tum += 1
        
            print(int(tum), end=" ")
        else:
            lceil = int(ceil(l / ab)) * ab
            rfloor = int(floor(r / ab)) * ab

            ans = ((rfloor - lceil) / ab) * num
            if l != lceil:
                ans += (num - pat[(l % ab) - 1])
            
            ans += (pat[r % ab])
            
            print(int(ans), end=" ")
    print()

            