Problem:
Score : 100 points

Problem Statement

You are parking at a parking lot. You can choose from the following two fee plans:

Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours.

Plan 2: The fee will be B yen, regardless of the duration.

Find the minimum fee when you park for N hours.

Constraints

1≤N≤20

1≤A≤100

1≤B≤2000

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

When the minimum fee is x yen, print the value of x.

Sample Input 1

7 17 120

Sample Output 1

119

If you choose Plan 1, the fee will be 7×17=119 yen.

If you choose Plan 2, the fee will be 120 yen.

Thus, the minimum fee is 119 yen.

Sample Input 2

5 20 100

Sample Output 2

100

The fee might be the same in the two plans.

Sample Input 3

6 18 100

Sample Output 3

100

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
public class Main{
    public static void main(String[] args) throws IOException{
        Reader r = new Reader();  
        int[] a = r.ii();
        int p1 = a[0]*a[1];
        int p2 = a[2];
        println(Math.min(p1, p2));
    }
    static void print(Object o){
        System.out.print(o.toString());
    }
    static void println(Object o){
        System.out.println(o.toString());
    }
    static int Int(String s){
        return Integer.parseInt(s);
    }
    static long Long(String s){
        return Long.parseLong(s);
    }
    static class Reader extends BufferedReader{
        Reader(){
            super(new InputStreamReader(System.in));
        }
        String s() throws IOException{
            return readLine();
        }
        String[] ss() throws IOException{
            return s().split(" ");
        }
        int i() throws IOException{
            return Int(s());
        }
        int[] ii() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            int[] ii = new int[size];
            for(int j = 0; j < size; j++) ii[j] = Int(ss[j]);
            return ii;
        }
        long l() throws IOException{
            return Long(s());
        }
        long[] ll() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            long[] ll = new long[size];
            for(int j = 0; j < size; j++) ll[j] = Long(ss[j]);            
            return ll;
        }
    }
}