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 sys
# sys.stdin = open('input.txt', 'r') 
# sys.stdout = open('output.txt', 'w')
 
import math
import collections
from sys import *
T = int(stdin.readline())
for _ in range(T):
    a,b,q = list(map(int,stdin.readline().split()))
    gc = math.gcd(a,b)
    lc = (a*b)//gc
    c = [0]*(lc+1)
    co = 0
    for i in range(1,lc+1,1):
        if((i%a)%b != (i%b)%a):
            co += 1
            c[i] = co
        else:
            c[i] = co
    c[-1] = co
    # print(c)
    for g in range(q):
        l,r = list(map(int,stdin.readline().split()))
        rsum = (r//lc)*co + c[r%lc]
        lsum = ((l-1)//lc)*co + c[(l-1)%lc]
        print(rsum-lsum,end= ' ')
    print('')