Problem:
Score : 200 points

Problem Statement

Snuke signed up for a new website which holds programming competitions.
He worried that he might forget his password, and he took notes of it.
Since directly recording his password would cause him trouble if stolen,
he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.

You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order.
Restore the original password.

Constraints

O and E consists of lowercase English letters (a - z).

1 \leq |O|,|E| \leq 50

|O| - |E| is either 0 or 1.

Input

Input is given from Standard Input in the following format:

O
E

Output

Print the original password.

Sample Input 1

xyz
abc

Sample Output 1

xaybzc

The original password is xaybzc. Extracting the characters at the odd-numbered positions results in xyz, and extracting the characters at the even-numbered positions results in abc.

Sample Input 2

atcoderbeginnercontest
atcoderregularcontest

Sample Output 2

aattccooddeerrbreeggiunlnaerrccoonntteesstt

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//  cout << char(i+48) << endl;//なぜかaは47と得る時がある。+48で出力もaにできる。
//  cout << char(97) << endl;//アスキーコードでaを出力
// sort(b.begin(), b.end());//bという配列を小さい方からソート
// reverse(b.begin(), b.end());//bという配列をリターン
// 10桁でdouble出力するパターン１ 
// printf("%.10f\n", ma);
// 10桁でdouble出力するパターン２
//cout << fixed << setprecision(10);
//  cout << ma << endl;

/*01 02 03 12 13 23　と６回見ていくパターン 
for(int i=0;i<n-1;i++){
  for(int j=i+1;j<n;j++){
    }
  }
*///vector<vector<int>> data(3, vector<int>(4));//int型の2次元配列(3×4要素の)の宣言
//int64_t a;
//10のi乗pow(10, i);
/*string s; stringでの文字列を数字型に変える方法
  cin >> s;
  rep(i,s.size()-2) {
  int a= (s.at(i)-'0')*100 + (s.at(i+1)-'0')*10+ s.at(i+2) -'0';
    */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
int main() {
  string x;
  string y;
  cin >> x >> y;
  if(x.size()==y.size()){
    for(int i=0;i<x.size();i++){
      cout << x.at(i) <<y.at(i);
    }
    cout<< endl;
  }
  else{
      for(int i=0;i<x.size()-1;i++){
      cout << x.at(i) <<y.at(i);
    }
    cout << x.at(x.size()-1) << endl;
  }
}