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 math
# import sys
# input=sys.stdin.buffer.readline

def lcm(a,b):
    return (a*b)//math.gcd(a,b)

def solve(r,y,b):
    if r == 0:
        return 0
    else:
        # y is the lcm
        div=r//y
        tot=r
        prod=div*y
        if div != 0 :
            tot=tot-(min(b,r-prod+1))
            tot=tot-(div-1)*(b)
        tot=tot-min(r,b-1)
        return tot


t=int(input())
for _ in range(t):
    a,b,q=[int(c) for c in input().split()]
    a,b=min(a,b),max(a,b)
    for i in range(q):
        l,r=[int(c) for c in input().split()]
        lc=lcm(a,b)
        if lc == b or r < b:
            print(0,end=" ")
        else:
            ans=solve(r,lc,b)
            if l > 1:
                ans-=solve(l-1,lc,b)
            print(ans,end=" ")
    print()
                        