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 gcd,ceil
import sys
input=sys.stdin.readline
from collections import defaultdict as dd
t=int(input())
while t:
    a,b,q=map(int,input().split())
    lcm=(a*b)//gcd(a,b)
    le=max(a,b)
    re=lcm-1
    m=max(a,b)
    #print(le,re)
    for i in range(q):
        l,r=map(int,input().split())
        li=-1
        ri=-1
        print(l,r)
        for i in range(l,r+1):
            if((i%a)%b!=((i%b)%a)):
                li=i
                break
        for i in range(li,r+1):
            if((i%a)%b==((i%b)%a)):
                ri=i-1
                break
        if(li==-1):
            print(0,end=" ")
            continue
        res=ri-li+1
        #print(res,li,ri)
        li=ri+m
        if(li>r):
            print(res,end=" ")
            continue
        
        ri=min(r,re*ceil(r/re))
        if((ri-li+1)//(m+re-le+1)==0):
            res+=(ri-li+1)
        else:
            res+=(re-le+1)*((ri-li+1)//(m+re-le+1))
            res+=max(0,r-ri+1)
        print(res,end=" ")
    print()
    t-=1
        