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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>

using namespace std;

#define FOR(i,a,b) for(int i = (a); i < (b); ++i)
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()

inline int toInt(string s) {int v; istringstream sin(s);sin>>v; return v;}

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector< vector<int> > Mat;
typedef tuple<int, int, int, int> T;

const int MOD = (int)1e9+7;
const int INF = (int)1e9;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int ddx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[] = {1, 1, 0, -1, -1, -1, 0, 1};

int main(){
    ios::sync_with_stdio(false);
	cin.tie(0);

    int N, A, B;
    cin >> N >> A >> B;

    cout << min(A*N, B) << endl;

    return 0;
}