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)

#pragma GCC optimize("O3")

#include "bits/stdc++.h"
#include "ext/pb_ds/tree_policy.hpp"
#include "ext/pb_ds/assoc_container.hpp"
#include "ext/rope"

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

typedef long long ll;
typedef pair<int, int> pii;
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for(int i = 0; i < (a); ++i)
#define pb push_back
#define mp make_pair
#define sz(x) (int)x.size()
#define f first
#define s second

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

struct custom_hash 
{
    static uint64_t splitmix64(uint64_t x) 
    {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const 
    {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

int main()
{
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n, a, b;
    cin >> n >> a >> b;
    cout << min(a * n, b);
    return 0;
}