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 __future__ import division, print_function
import sys
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

import os, sys, bisect, copy
from collections import defaultdict, Counter, deque
#from functools import lru_cache   #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------
from math import*
for _ in range(int(input())):
    a,b,q = mapi()
    lcm =  (a*b)//gcd(a,b)
    ans = []
    b = max(a,b)
    for qe in range(q):
        x,y = mapi()
        res = 0
        #res = max(y-max(x,b)+1,0)
        res = (y//lcm)*(lcm-b)
        rem = y%lcm
        if rem >=b:
            res+=(rem-b+1)
        x-=1
        tmp = (x//lcm)*(lcm-b)
        rem = x%lcm
        if rem >=b:
            tmp+=(rem-b+1)
        res-=tmp
        ans.append(res)
    print(*ans)
