Problem: 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 

Solution (in JAVA):

Solution:
import java.util.*;


public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        StringBuilder ans = new StringBuilder();
        int t = sc.nextInt();
        while (t-- > 0) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int q = sc.nextInt();
            int LCM = (int)lcm(a,b);
            int cnt = 0;
            int[] sum = new int[LCM];
            for (int x = 0; x < LCM; x++) {
                if (((x%a)%b)!=((x%b)%a)) {
                    cnt++;
                }
                sum[x] = cnt;
            }
            for (int i = 0; i < q; i++) {
                long l = sc.nextLong();
                long r = sc.nextLong();
                ans.append(count(r,LCM,sum)-count(l-1,LCM,sum)).append(' ');
            }
            ans.append('\n');
        }
        System.out.print(ans);
    }

    private static long count(long y, int lcm, int[] sum) {
        return y/lcm*sum[lcm-1]+sum[(int)(y%lcm)];
    }

    private static long lcm(long a, long b) {
        return a*b/gcd(a,b);
    }

    private static long gcd(long a, long b) {
        return b > 0 ? gcd(b,a%b) : a;
    }
}