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.io.*;
import java.math.BigInteger;
import java.util.*;


public class Main{
    static long [][] memo;
    static long [] pow = new long[33];
    static ArrayList <Integer> [] graph;
    static Integer [] d;
    static int [] dist;
    static boolean[][] visited;
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        // Start writing your solution here. -------------------------------------
        long MOD = 1000000007;
        int mod = 2019;
        int t = sc.nextInt();
        //int t = 1;
        while(t -- > 0)
        {
            int a = sc.nextInt(), b = sc.nextInt();
            int q = sc.nextInt();
            int GCD = gcd(a, b), LCP = a * b / GCD;
            long res[] = new long[LCP];
            for(int i = 1; i < LCP; i++)
            {
                res[i] = res[i - 1];
                int r1 = (i % a) % b, r2 = (i % b) % a;
                if(r1 != r2) res[i]++;
            }
            //System.out.println(Arrays.toString(res));
            for(int i = 0; i < q; i++)
            {
                long l = sc.nextLong(), r = sc.nextLong();
                long ans = solve(r, res) - solve(l - 1, res);
                out.print(ans + " ");
            }
            out.println("");
        }
        out.close();
    }
    public static int gcd(int a, int b)
    {
        if(a % b == 0) return b;
        return gcd(b, a % b);
    }
    public static long solve(long x, long[] res){
        int n = res.length;
        long a = x / n;
        int b = (int)(x % n);
        return res[n - 1] * a + res[b];
    }
    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    //--------------------------------------------------------
}