Problem: 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 

Solution (in PYTHON3):

Solution:
from sys import stdin
input=stdin.readline
import bisect
import math
#i = bisect.bisect_left(a, k)
#list=input().split(maxsplit=1)
for xoxo in range(1):
    #a=[]
    for _ in range (int(input())):
    #n=int(input()) 
    #ans=[]
        a,b,q=map(int, input().split())
        g=math.gcd(a,b)
        a,b=min(a,b),max(a,b)
        ans=0
        if a==b:
            for i in range(q):
                l,r=map(int, input().split())
            ans=[0]*q
            print(*ans)
        else:
            k=(a*b)//g
            
            a=[]
            for i in range(q):
                ans=0
                l,r=map(int, input().split())
                t=l-r+1
                ll=(l//k)*k
                rr=(r//k)*k
                ans=0
                if ll==rr:
                    ans+=(max(0,(r-max(l-1,(ll+b-1)))))
                    a.append(ans)
                else:
                    l=max(l,ll+b)
                    if rr+b>r:
                        r=rr-1
                    t=r-l+1
                    #print(l,r,((((l-1)//k)-(r//k))*b))
                    ans+=(t+((((l-1)//k)-(r//k))*b))
                    a.append(ans)
            print(*a)
    #a=list(map(int, input().split()))a
    #ans=0