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 collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math
import random

# sys.stdin = open('input')

def inp():
    return map(int, raw_input().split())

def inst():
    return raw_input().strip()

def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x

def my_main():
        t = inp()[0]
        for _ in range(t):
            a, b, q = inp()
            if b>a:
                a, b = b, a
            lcm = a*b / gcd(a, b)
            # print lcm
            pr = []
            for _ in range(q):
                l, r = inp()
                ll = l/lcm
                rr = r/lcm

                ans = (rr-ll+1)*(a)
                for i in range(ll*lcm, ll*lcm+a):
                    if i<l:
                        ans -= 1
                for i in range(rr*lcm, rr*lcm+a):
                    if i>r:
                        ans -= 1
                pr.append(r-l+1-ans)
            print ' '.join(map(str, pr))

        # for i in range(30):
        #     if (i%4)%6 == (i%6)%4:
        #         print i





my_main()
