Problem:
Score : 100 points

Problem Statement

Given are two strings S and T consisting of lowercase English letters. Concatenate T and S in this order, without space in between, and print the resulting string.

Constraints

S and T are strings consisting of lowercase English letters.

The lengths of S and T are between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

S T

Output

Print the resulting string.

Sample Input 1

oder atc

Sample Output 1

atcoder

When S = oder and T = atc, concatenating T and S in this order results in atcoder.

Sample Input 2

humu humu

Sample Output 2

humuhumu

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for (int (i) = 0; (i) < (int)(n); i++) 
#define FOR(i, a, b) for(int (i) = a; (i) < (int)b; i++)
#define RREP(i, n) for(int (i)=((int)(n)-1); (i)>=0; i--)
#define RFOR(i, a, b) for(int (i) =((int)(b)-1); (i)>=(int)a; i--)
#define ALL(v) (v).begin(),(v).end()
#define MOD 1000000007
#define NIL -1
#define FI first
#define SE second
#define MP make_pair
#define PB push_back
#define SZ(x)  (int)x.size()
#define SP(x)  setprecision((int)x)


using namespace std ;
typedef long long ll;
typedef vector<int> vint;
typedef vector<vint> vvint;
typedef vector<string> vstr;
typedef pair<int, int> pii;
const int INF = 1e9;
const ll LINF = 1e18;
const double EPS = 1e-9;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} //最小公倍数

//-------------------------------------------------
// 
void yes(){
  cout <<"Yes"<<endl ;
}
void no(){
  cout <<"No"<<endl ;
}


//-------------------------------------------------
// メモ
/*






*/
//-------------------------------------------------


int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  string s,t ;
  cin >> s >> t ;
  cout << s << t <<endl ;
  
  
  
  return 0 ;
}