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 java.io.*;
import java.util.*;
import java.lang.*;

public class Rextester{
    static int gcd(int a,int b){
        if(a%b==0){
            return b;
        }
        return gcd(b,a%b);
    }
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        StringBuffer sb = new StringBuffer();
        int z=0;
        while(t-->0){
            StringTokenizer st = new StringTokenizer(br.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            int q = Integer.parseInt(st.nextToken());
            int lcm = a*b;
            lcm/=gcd(a,b);
            while(q-->0){
                z++;
                StringTokenizer st1 = new StringTokenizer(br.readLine());
                long l = Long.parseLong(st1.nextToken());
                long r = Long.parseLong(st1.nextToken());
                long lbeg = (long)lcm*(l/lcm+1);
                long rend=(long)lcm*(r/lcm); 
                if(lbeg>=rend){
                    long ans = 0;
                    for(long i=l;i<=r;i++){
                        if((i%a)%b==(i%b)%a){
                            ans++;
                        }
                    }
                    sb.append(r-l+1-ans).append(" ");
                    continue;
                }        
                long count = (rend-lbeg)/lcm;
                long ans = count*Math.max(a,b);
                if(lbeg-lcm+Math.max(a,b)>=l){
                    ans+=lbeg-lcm+Math.max(a,b)-l;
                }
                if(rend+Math.max(a,b)>=r){
                    ans+=r-rend+1;
                }
                else{
                    ans+=Math.max(a,b);
                }
                if(z==7033 && r-l+1-ans==205){
                    System.out.println(a+" "+b+" "+l+" "+r);
                }
                sb.append(r-l+1-ans).append(" ");
            }
            sb.append("\n");
        }
        br.close();
        System.out.println(sb);
    }
}
            