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 
import sys
input = sys.stdin.readline
def getInt(): return int(input())
def getVars(): return map(int, input().split())
def getList(): return list(map(int, input().split()))
def getStr(): return input().strip()
## -------------------------------

def nokAB(a,b):
    m = a*b
    while a != 0 and b != 0:
        if a > b:
            a %= b
        else:
            b %= a
    return m // (a+b)

tt = getInt()
for _ in range(tt):
    a, b, q = getVars()
    nok = nokAB(a, b)
    arr = []
    for i in range(q):
        l, r = getVars()
        res = max((l//nok) * nok + max(a, b) - l, 0)
        res += (r // nok - l // nok) * max(a, b)
        res -= max((r // nok) * nok + max(a, b) - r - 1, 0)
        arr.append(r - l + 1 - res)
    print(*arr)
            

        
                        
                    
                
    
