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 ceil

def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a%b)
t = int(input())
for _ in range(t):
    a, b, q = map(int, stdin.readline().split())

    if max(a, b)%min(a, b) == 0:
        for _ in range(q):
            l, r = map(int, stdin.readline().split())
            print(0)

    else:
        lcm = (a*b)//gcd(a, b)
        m = max(a, b)
        for _ in range(q):
            l, r = map(int, stdin.readline().split())
            ans = 0
            totnum = r - l + 1
            if r < m:
                print(0)
                continue

            if l%lcm >= m:
                start = ceil(l/lcm)
            else:
                ans += (m - l%lcm)
                start = ceil(l/lcm)

            if start>r:
                print(0)
                continue
            else:
                print(m, lcm, start, r)
                ans += (m*((r-m)//(start*lcm)))

            # ans += m*((r-start-m)//lcm)
            print(totnum - ans)
