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 as mt 
import sys,string
input=sys.stdin.readline
#print=sys.stdout.write
#import random
from heapq import heappush,heapify,heappop
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
def lcm(a,b):
    return (a*b)//mt.gcd(a,b)
t=I()
for _ in range(t):
    a,b,q=M()
    x=lcm(a,b)
    for i in range(q):
        l,r=M()
        d=r-l+1
        ans1=(r//x)*max(a,b)
        
        r%=x
        ans1+=min(max(a,b)-1,r)
        l-=1
        ans2=(l//x)*max(a,b)
        l%=x
        ans2+=min(max(a,b)-1,l)
        equal=ans1-ans2
        print(d-equal,end=" ")
    print()
    
