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: Wrong Answer)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

const int N = 1e5 + 10 , M = 5e5+10 ,  MOD = 1e9+7;
const int OO = 0x3f3f3f3f;
const double EPS = 1e-3;

#define clr(arr,val) memset(arr , val , sizeof (arr))
#define loop(i,n) for(int i = 0;i < int(n);i++)
#define rloop(i,n) for(int i = int(n)-1;i >= 0;i--)
#define range(i,a,b) for(int i = int(a);i <= int(b);i++)
#define ALL(v) ((v).begin()) , ((v).end())
#define SZ(v)  ((int)((v).size()))
typedef pair <int,int> pii ;
typedef vector < pii > vii ;
typedef vector < int > vi;
int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction
//int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 direction
//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction
//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
 


int a,b;



int main() {

    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
	
    string ans = "";

    cin >> a >> b;
    if(a < b){
        for(int i = 0 ; i < b ; ++i)
            ans += to_string(b);
        cout << ans << "\n";

    }
    else if(a > b){
        for(int i = 0 ; i < a ; ++i)
            ans += to_string(b);
        cout << ans << "\n";

    }
    else {
        for(int i = 0 ; i < a ; ++i)
            ans += to_string(b);
        cout << ans << "\n";
    }


    
    return 0;
}