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.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            long a=sc.nextInt();
            long b=sc.nextInt();
            long q=sc.nextInt();
            long min=(a<b)?a:b;
            long max=(a>b)?a:b;
            if(max%min==0){
                for (int i=0;i<q;i++) {
                    long l=sc.nextLong();
                    long r=sc.nextLong();
                    System.out.print(0+" ");
                }
                System.out.println();
                continue;
            }
            a=min;
            b=max;
            long blength=b*(a-1);
            long lcm=a*b;
            while(q-->0){
                long sum=0;
                long l=sc.nextLong();
                long r=sc.nextLong();
                if(r<b) {
                    System.out.print(0+" ");
                    continue;
                }
                long d1=l/lcm;
                long d2=r/lcm;
                sum+=((d2-d1)-1)*blength;
                //for l
                long lcm1=lcm*d1;
                long left=lcm1-1;
                long right=lcm1+b;
                if(l<=right)
                    sum+=blength;
                else{
                    right=(lcm*(d1+1))-1;
                    right=right-l;
                    right++;
                    sum+=right;
                }
                // for r
                lcm1=lcm*d2;
                left=lcm1+b;
                right=(lcm*(d2+1))-1;
                if(r==right)
                    sum+=blength;
                else if(left==r)
                    sum+=1;
                else if(r<right && r>left){
                    right=r-left;
                    right++;
                    sum+=right;
                }
                System.out.print(sum+" ");
            }
            System.out.println();
        }
    }
}