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


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

def hey(x,b,l,r):
    # x=l//h and r//h and l//h-1
    g=0
    for i in range(b):
        if l<=x+i<=r :
            g+=1
    return(g)


def find_ans(a, b, l, r):

    h = hcm(a, b)

    if a > b:
        a, b = b, a

    if b % a == 0 or r < b:
        return(0)

    ans = r-l+1
    ans = ans-max(b-l, 0)-max((r//h - l//h - 1)*b,0)
    if  r//h<=2 :
        i=max(l//h -1 ,1 ) 
        while True:
            k=h*i
            for j in range(b):
                if l<=k+j<=r :
                    ans-=1
                elif k+j>r :
                    return(ans)
            i+=1 
    ans-=(hey((l//h)*h,b,l,r)+hey((l//h-1)*h,b,l,r)+hey((r//h)*h,b,l,r))
    return(ans)    

for _ in range(int(input())):
    a, b, t = map(int, input().split())
    for ll in range(t):
        l, r = map(int, input().split())
        print(find_ans(a, b, l, r), end=' ')
    print()
