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
for ii in range(int(input())):
    t = [int(i) for i in input().split()]
    a = max(t[0],t[1])
    b = min(t[0],t[1])
    q = t[2]
    size = 1 
    arr = [0 for i in range(a*b)]
    y = a*b
    arr[0] = 1
    for i in range(1,a*b):
        if (i%a)%b == (i%b)%a:
            size += 1
        arr[i] = size
    ans = []
    for i in range(q):
        x = [int(i)for i in input().split()]
        l = x[0]
        r = x[1]
        #print((r//y)*size+arr[(r+1)%y],'and',((l-1)//y)*size+arr[(l)%y])
        ans.append((r-l+1-((r//y)*size+arr[(r)%y]-((l-1)//y)*size-arr[(l-1)%y])))
    print(*ans)

