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.*;

public class YetAnotherCountingProblemCF {
    static boolean[] condn = new boolean[40001];
    static int[] count = new int[40001];
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- != 0) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int q = sc.nextInt();
            Arrays.fill(condn, false);
            Arrays.fill(count, 0);
            preComp(a, b);
            while(q-- != 0) {
                long l = sc.nextLong();
                long r = sc.nextLong();
                System.out.print(solve(l, r, a, b) + " ");
            }
            System.out.println();
        }
    }    

    static void preComp(int a, int b) {
        int cnt = 0;
        for(int i=0; i<=(a*b); i++) {
            int c1 = (i%a)%b;
            int c2 = (i%b)%a;
            if(c1 != c2) {
                condn[i] = true;
                cnt++;
            }
            count[i] = cnt;
        }
    }

    static long solve(long l, long r, int a, int b) {
        long cntr = count(r, a, b);
        long cntl = count(l-1, a, b);
        return cntr - cntl;
    }
    //Count from 0 to r
    static long count(long r, int a, int b) {
        int x = a * b;
        long ans = (long)(r/x) * (long)count[x-1];
        ans += count[(int)(r%x + x)%x];
        return ans;
    }
}
