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
LI=lambda:list(map(int, sys.stdin.readline().split()))
MI=lambda:map(int, sys.stdin.readline().split())
SI=lambda:sys.stdin.readline().strip('\n')
II=lambda:int(sys.stdin.readline())

from math import gcd
for _ in range(II()):
    a, b, q=MI()
    lcm=a*b//gcd(a, b)
    m=max(a, b)
    res=[]
    for __ in range(q):
        l, r=MI()
        n1, n2=l//lcm, r//lcm
        cnt=0
        for i in range(lcm*n1, lcm*n1+m):
            if l<=i<=r:
                cnt+=1
        #print(cnt)
        #print(cnt)
        if n2!=n1:
            for i in range(lcm*n2, lcm*n2+m):
                if l<=i<=r:
                    cnt+=1
            cnt+=(n2-n1-1)*m
        #print(cnt)
        res.append(r-l+1-cnt)
    print(*res)
