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 (GCC 5.4.1) (judge verdict: Accepted)


/*
LANG:C++
PROB:

*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
#include <iostream>
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)


using namespace __gnu_pbds;
using namespace std;


typedef tree<int, null_type, less<int>, rb_tree_tag,
            tree_order_statistics_node_update>
    ordered_set;

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;
typedef unsigned long long ui64;

template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }
const int mod=1000000007;
  i64 poww(i64 a, i64 b)
  {
    a%=mod;
    i64 res=1;
    while(b)
      {
if(b&1)res=res*a%mod;

      a=a*a%mod;
      b/=2;

      }
      return res;
  }
  i64 inv(i64 a)
  {
    return poww(a,mod-2);
  }


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

    // ifstream fin(".in");
    // ofstream fout(".in");
int a,b;
cin>>a>>b;
if(b>a)
cout<<string(b,'0'+a);
else
cout<<string(a,'0'+b);




    return 0;
  }