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 gcd, ceil, floor

t = int(input())
for _ in range(t):
    a, b, q = list(map(int, input().split()))
    if a > b:
        a, b = b, a
    lcm = (a * b) // gcd(a, b)
    ans = []
    for _ in range(q):
        l, r = list(map(int, input().split()))
        if l < b :
            l = b
        if r < b:
            ans.append(0)
        elif a == b:
            ans.append(0)
        elif lcm > r :
            ans.append(r - l + 1)
        else:

            # print(l,b)
            start = int(ceil(l / lcm)) * lcm
            # print(start)
            if start > r:
                ans.append(r - l + 1)
                continue
            if start + b >= r:
                ans.append(r - l + 1 - (r - start+1))
                continue

            temp = int(floor((r - start) / lcm))

            x = temp * b
            # print(x)
            if temp != 0:
                if start + lcm * temp + b <= r:
                    x += b
                else:
                    # print("asdasdsa")
                    x = x + (r - (start + lcm * temp) + 1)
                    # print(x,)
            else:
                x += b
            ans.append(r - l + 1 - x)
    print(*ans)
