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 
from sys import stdin, stdout
from math import ceil
def gcd(a,b): 
    if a == 0: 
        return b 
    return gcd(b % a, a) 

def find(A,B,Q,L,R):
    if A%B==0 or B%A==0: return 0
    A,B=sorted([A,B])
    K=A*B//gcd(A,B)
    R+=1
    Z=max(0,(R//K-L//K-1)*B)
    # if (R-L+1)%K:
    x=L-(L//K)*K
    Z+=max(0,min(B-x,R-L))
    y=R-(R//K)*K
    if R//K!=L//K:
        Z+=min(y,B)
    
    return R-L-Z
# def find2(A,B,Q,L,R):
#     c=0
#     for i in range(L,R+1):
#         if i%B%A!=i%A%B: c+=1
#     return c
# for x in range(1,100):
#     for y in range(1,100):
#         for i in range(1,200):
#             for j in range(i,200):
#                 if find2(x,y,5,i,j)!=find(x,y,5,i,j):
#                     print(i,j)
# print(find2(4,2,5,3,4))
# print(find(4,2,5,3,4))
# find(4,2,5,i,j):
def main():
    for _ in range(int(stdin.readline())):
        A,B,Q=list(map(int, stdin.readline().split()))
        arr=[]
        for __ in range(Q):
            L,R=list(map(int, stdin.readline().split()))
            arr.append(find(A,B,Q,L,R))
        print(' '.join(map(str,arr)))
main()