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)

#include<bits/stdc++.h>
//ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
ll mod=1000000007;
ll mod2=998244353;
ll mod3=1000003;
ll mod4=998244853;
ll inf=1000000000000000000;
long double pi=3.14159265358979323846L;
double eps=1e-12;
#define rep(i,m,n) for(int i=m;i<n;i++)
#define rrep(i,n,m) for(int i=n;i>=m;i--)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll lmax(ll a,ll b){
    if(a<b)return b;
    else return a;
}
ll lmin(ll a,ll b){
    if(a<b)return a;
    else return b;
}
ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(b==0)return a;
    if(a%b==0)return b;
    return gcd(b,a%b);
}
ll Pow(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k&1)ret*=now;
        now*=now;
        k/=2;
    }
    return ret;
}
ll gya[1000010];
ll kai[1000010];
ll beki(ll n,ll k,ll md){
  ll ret=1;
  ll now=n;
  while(k>0){
    if(k%2==1){
      ret*=now;
      ret%=md;
    }
    now*=now;
    now%=md;
    k/=2;
  }
  return ret;
}
ll gyaku(ll n,ll md){
  return beki(n,md-2,md);
}

void nckinit(ll n,ll md){
  kai[0]=1;
  kai[1]=1;
  for(int i=2;i<=n;i++){
    kai[i]=kai[i-1]*i;
    kai[i]%=md;
  }
  gya[n]=gyaku(kai[n],md);
  for(int i=n-1;i>=1;i--){
    gya[i]=gya[i+1]*(i+1);
    gya[i]%=md;
  }
  gya[0]=1;
}
ll nck(ll n,ll k,ll md){
  if(k<0)return 0;
  if(k==0||n==k)return 1;
  ll ret=kai[n];
  ret*=gya[n-k];
  ret%=md;
  ret*=gya[k];
  ret%=md;
  return ret;
}
ll npk(ll n,ll k,ll md){
  if(k<0)return 0;
  if(k==0)return 1;
  ll ret=kai[n];
  ret*=gya[n-k];
  ret%=md;
  return ret;
}
int main(){
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int a,b;cin>>a>>b;
    if(a<b){
        rep(i,0,b)cout<<a;
        cout<<endl;
    }
    else{
        rep(i,0,a)cout<<b;
        cout<<endl;
    }
}