Problem:
Score : 200 points

Problem Statement

Given are 1-digit positive integers a and b. Consider these two strings: the concatenation of b copies of the digit a, and the concatenation of a copies of the digit b. Which of these is lexicographically smaller?

Constraints

1 \leq a \leq 9

1 \leq b \leq 9

a and b are integers.

Input

Input is given from Standard Input in the following format:

a b

Output

Print the lexicographically smaller of the two strings. (If the two strings are equal, print one of them.)

Sample Input 1

4 3

Sample Output 1

3333

We have two strings 444 and 3333. Between them, 3333 is the lexicographically smaller.

Sample Input 2

7 7

Sample Output 2

7777777

Solution in C++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#include <climits>
#include <cstring>

#define rep(i, m, n) for(int i=int(m);i<int(n);i++)
#define all(c) begin(c),end(c)

template<typename T1, typename T2>
inline void chmin(T1 &a, T2 b) { if (a > b) a = b; }

template<typename T1, typename T2>
inline void chmax(T1 &a, T2 b) { if (a < b) a = b; }

typedef long long int ll;
using ll = long long int;
using ull = long long unsigned int;
using Int = long long int;
using namespace std;
#define INF (1 << 30) - 1
#define INFl (ll)5e15
#define DEBUG 0
#define dump(x)  cerr << #x << " = " << (x) << endl
#define MOD 1000000007


//edit
class Solve {
public:
    void solve() {
        int a, b;
        cin >> a >> b;

        string s, t;
        for (int i = 0; i < b; ++i) {
            s += char(a + '0');
        }
        for (int i = 0; i < a; ++i) {
            t += char(b + '0');
        }

        if (s < t) {
            cout << s << endl;
        } else {
            cout << t << endl;
        }

    }
};


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    Solve().solve();


    return 0;
}