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.Arrays;
import java.util.Scanner;

public class Main {
    // https://codeforces.com/problemset/problem/1342/C
    // creating flag array , 1 where condition satisfies
    // length = 40000 bcz 1<=a,b <=200
    public static int[] flag = new int[40005];
    public static int a;
    public static int b;

    public static void create_flag(){
        flag[0] = 0;
        //1->ab-1...ab->2ab-1
        //accumulative sum array
        for(int i = 1;i<a*b;i++){
            if((i%a)%b != (i%b)%a){
                flag[i] = flag[i-1]+1;
            }
            else{
                flag[i] = flag[i-1];
            }
        }
    }
    public static long countNoOfX(long n){
        if(n < a*b){
            return flag[(int)n];
        }
        long count = (n/(a*b))*flag[a*b-1];
        count += flag[(int)(n%(a*b))];
        return count;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            a = sc.nextInt();
            b = sc.nextInt();
            int q = sc.nextInt();
            // (x % a) % b = ((ab+x) % a) % b => x%a = (ab+x)%a
            // (x % b) % a = ((ab+x) % b) % a
            // values repeat after ab
            // create flag array till ab
            create_flag();
            while (q-- > 0) {
                long l = sc.nextLong();
                long r = sc.nextLong();
                System.out.print(countNoOfX(r)-countNoOfX(l-1)+" ");
            }
            System.out.println();
            Arrays.fill(flag,0);
        }
        sc.close();
    }
}