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,sys
input=sys.stdin.readline
T=int(input())
for _ in range(T):
    a,b,q=map(int,input().split())
    if (b<a):
        a,b=b,a
    c=(a*b)//math.gcd(a,b)
    A=[0 for i in range(c)]
    #print(c)
    for i in range(c):
        if (i>=b):
            A[i]=A[i-1]+1
    B=[]
    for i in range(q):
        v=0
        l1,r1=map(int,input().split())
        l=max(l1%c,1)
        r=r1%c
        l2=l1//c
        r2=r1//c
        vc=A[r]-A[l-1]+(A[c-1]*(r2-l2))
        #print(vc)
        B.append(vc)
    print(*B)
    
