Problem: 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 

Solution (in PYTHON3):

Solution:
import io,os
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
from itertools import accumulate
def gcd(x,y):
    if x<y:
        x,y=y,x
    if y==0:
        return x
    return gcd(y,x%y)
def lcm(x,y):
    return x*y//gcd(x,y)

def main():
    t=int(input())
    for _ in range(t):
        a,b,q=map(int,input().split())
        lcm_ab=lcm(a,b)
        Cycle=[None]*lcm_ab
        for i in range(lcm_ab):
            Cycle[i]=int((i%a)%b!=((i%b)%a))
        Acc=[0]+list(accumulate(Cycle))
        Ans=[]
        for _ in range(q):
            l,r=map(int,input().split())
            l_cycle=l//lcm_ab
            l_idx=l%lcm_ab
            r_cycle=r//lcm_ab
            r_idx=r%lcm_ab
            ans=Acc[-1]*(r_cycle-l_cycle)+Acc[r_idx+1]-Acc[l_idx]
            Ans.append(ans)
        sys.stdout.write(' '.join(map(str,Ans))+'\n')
    
if __name__=='__main__':
    main()