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 sys
import math
t=int(sys.stdin.readline())
for _ in range(t):
    x,y,q=map(int,sys.stdin.readline().split())
    a=min(x,y)
    b=max(x,y)
    lcm=(a*b)//math.gcd(a,b)
    #print(lcm,'lcm')
    res=[]
    for i in range(q):
        l,r=map(int,sys.stdin.readline().split())
        
        A=l//lcm
        B=l%lcm
        left=(A)*b + min(b-1,B)+1
        C=r//lcm
        D=r%lcm
        right=(C)*b + min(b-1,D) + 1
        #print(l,'l',r,'r',A,'A',B,'B',C,'C',D,'D')
        left-=1
        right-=1
        #print(A,'A',B,'B')
        #print(left,'left')
        #print(C,'C',D,'D')
        #print(right,'right')
        #print(left,'left',right,'right')
        
        if 0<=l%lcm<=b-1:
            ans=right - left + 1
        else:
            ans = right - left
        ans=(r-l+1-ans)
        res.append(ans)
    print(*res)
