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
from collections import *
from bisect import *
from math import *
input=sys.stdin.readline
t=int(input())
while(t):
    t-=1
    a,b,q=map(int,input().split())
    fac=(a*b)//gcd(a,b)
    rrr=[]
    for i in range(q):
        s1=0
        s2=0
        l,r=map(int,input().split())
        s=max(a,b)
        if(l<s):
            s1+=l
        else:
            s1+=s-1
            q1=max(0,(l//(fac)))
            if(q1!=0):
                s1+=max(0,(q1-1))*(s)
            if(q1!=0):
                if(l%(fac)==0):
                    s1+=s
                else:
                    s1+=min(s,(l%(fac))+1)
                    
        if(r<s):
            s2+=r
        else:
            s2+=s-1
            q2=max(0,(r//(fac)))
            if(q2!=0):
                s2+=max(0,q2-1)*(s)
            if(q2!=0):
                if(r%(fac)==0):
                    s2+=s
                else:
                    s2+=min(s,(r%(fac))+1)
        #print(s1,s2)
        s2=s2-s1
        #print(s2)
        if(((l%a)%b)==((l%b)%a)):
            zz=1
        else:
            zz=0
        rrr.append(r-l-s2+1-zz)
        #print(rrr[-1])
    print(*rrr)
        
            
