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 countDivisibles(A, B, M): 
      
    # Add 1 explicitly as A is divisible by M 
    if (A % M == 0): 
        return ((B // M) - (A // M)) + 1
  
    # A is not divisible by M 
    return ((B // M) - (A // M))  
def lcm(A, B):
    lc=(A*B)//math.gcd(A,B)
    return lc
for i in range(int(input())):
    a,b,q=map(int,input().split())
    lc=lcm(a,b)
    for j in range(q):
        l,r=map(int,input().split())
        e=(r//lc)*lc
        e1=(l//lc)*lc
        tm=0
        ta=0
        if e+max(a,b)-1>r:
            tm=e+max(a,b)-r-1
        if e1+max(a,b)-1>=l:
            ta=e1+max(a,b)-l
        rw=countDivisibles(l, r, lc)*max(a,b)
        #print(rw)
        rw=rw-tm+ta
        print(max(0,r-l+1-rw))
        
        