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 os import path
if(path.exists('C:/Users/prana/Desktop/sublime/input.txt')):
    sys.stdin = open("C:/Users/prana/Desktop/sublime/input.txt","r")
    sys.stdout = open("C:/Users/prana/Desktop/sublime/output.txt","w")

# ----------------------------------------------------------------------
import math

def solve():

    abq=list(map(int,input().split()))
    a=abq[0]
    b=abq[1]
    q=abq[2]
    if a>b:
        a^=b
        b^=a
        a^=b

    an=[]
    for i in range(q):
        lr=list(map(int,input().split()))
        l=lr[0]
        r=lr[1]
        # for i in range(l,r+1):
        #     print(i,((i%a)%b),((i%b)%a),file = sys.stderr)
        lc=(a*b)//math.gcd(a,b)
        print(a,b,lc,file = sys.stderr)
        ans=max(b,r)
        if(b==lc):
            an.append(0)
        else:
            if(ans==b and b!=r):
                an.append(0)
            else:
                p=max(b,l)
                
                rm=r%lc
                
                lm=p%lc
                
                
                ch=(r//lc-p//lc)
                temp1=(ch)*(b)
                templ=max(0,b-lm)
                tempr=max(1,b-rm)-1
                temp=temp1-tempr+templ
                n=r-p+1
                print(temp1,templ,tempr,file = sys.stderr)
    
                
                ans=n-temp
               
                an.append(ans)
                print(lm,rm,ch,r-p,temp1,file = sys.stderr)
                

    for i in an:
        print(i,end=" ")
    print(end="\n")

T = 1
T = int(input())
for _ in range(1,T+1):
    print("Case #", _ , file = sys.stderr)
    solve()
                
    