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 math import *
from bisect import bisect_left, bisect_right
from queue import Queue
from sys import stdin, stdout
from collections import Counter
input = lambda: stdin.readline().strip()
fast_print = stdout.write

t = int(input())
for _ in range(t):
    a, b, q = map(int, input().split())
    prefix = []
    cnt = 0
    for i in range(a*b):
        if (i%b)%a!=(i%a)%b:
            cnt+=1
        prefix.append(cnt)
    for i in range(q):
        l, r = map(int, input().split())
        minicnt = 0
        if l!=0:
            minicnt+=cnt-prefix[(l%(a*b))-1]
        else:
            minicnt+=cnt
        if r!=a*b-1:
            minicnt+=prefix[r%(a*b)]
        else:
            minicnt+=cnt
        cnt_back = (a*b-(l%(a*b)))
        cnt_front = (r%(a*b))+1
        if r-l+1<a*b:
            if r-l+1==cnt_front+cnt_back:
                fast_print(str(minicnt)+' ')
            else:
                fast_print(str(prefix[r%(a*b)]-prefix[(l%(a*b))-1])+' ')
        else:
            leftover = r-l+1-(cnt_front+cnt_back)
            fast_print(str(minicnt+cnt*(leftover//(a*b)))+' ')
    fast_print('\n')
