Problem:
Score : 100 points

Problem Statement

N of us are going on a trip, by train or taxi.

The train will cost each of us A yen (the currency of Japan).

The taxi will cost us a total of B yen.

How much is our minimum total travel expense?

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq A \leq 50

1 \leq B \leq 50

Input

Input is given from Standard Input in the following format:

N A B

Output

Print an integer representing the minimum total travel expense.

Sample Input 1

4 2 9

Sample Output 1

8

The train will cost us 4 \times 2 = 8 yen, and the taxi will cost us 9 yen, so the minimum total travel expense is 8 yen.

Sample Input 2

4 2 7

Sample Output 2

7

Sample Input 3

4 2 8

Sample Output 3

8

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

#include<iostream>
#include<cmath>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<iomanip>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> ipair;
bool lessPair(const ipair& l, const ipair& r){return l.second < r.second;}
bool morePair(const ipair& l, const ipair& r){return l.second > r.second;}

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

const ll MOD = 1e9 + 7;
// const long long INF = 1LL<<60;
void add(long long &a, long long b) { a += b; if (a >= MOD) a -= MOD; }
void sub(long long &a, long long b) { a -= b; if (a < 0) a += MOD; }
void mul(long long &a, long long b) { a *= b; a %= MOD; }
ll llmin(ll a, ll b) { if (a < b) return a; else return b; }
ll llmax(ll a, ll b) { if (a < b) return b; else return a; }
ll llabs(ll a) { if (a >= 0) return a; else return - a; }
ll llmodpow(ll a, ll n) {
   if (n == 0) return 1;
   ll tmp = llmodpow(a, n / 2);
   mul(tmp, tmp);
   if (n & 1) mul(tmp, a);
   return tmp;
}

int main() {
   int N, A, B;
   cin >> N >> A >> B;
   cout << min(A * N, B) << endl;
   return 0;
}