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)

#include <bits/stdc++.h>
#define ll long long
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;

//定数
//円周率
const double pi = 3.141592653589793238462643383279;
//天井
const int INF = 1000000000; // = 10^9
const ll LINF = 100000000000000000; // = 10^17
//ABC文字列
const string ABC = "ABCDEFGHIJKLMNOPQRSTUVWXYZABC";
const string abc = "abcdefghijklmnopqrstuvwxyzabc";
//よくあるmodくん
const ll MOD = 1000000007; // = 10^9 + 7

//データ構造
//隣接リスト用構造体（有向グラフ向け）
struct edge {
  ll to; // E.toでその辺の終点へアクセスできる
  ll cost; // e.costでその辺の重みにアクセスできる
};
//Union_Find木
struct UnionFind {
  vector<int> UF; // UF.at(i) : iの親の番号
  vector<int> SIZE; // SIZE.at(root(i)) : iと連結されてる要素の数
  UnionFind(int N) : UF(N), SIZE(N, 1) { // 最初は全てが根であるとして初期化
    for(int i = 0; i < N; i++) {
      UF.at(i) = i;
    }
  }
  int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
    if (UF.at(x) == x) {
      return x;
    }
    return UF.at(x) = root(UF.at(x));
  }
  void unite(int x, int y) { // xとyの木を併合
    int rx = root(x); // xの根をrx
    int ry = root(y); // yの根をry
    if (rx == ry) {
      return; // xとyの根が同じ(=同じ木にある)時はそのまま
    }
    // xとyの根が同じでない(=同じ木にない)時：小さい方の根を大きい方の根につける
    if (SIZE.at(rx) < SIZE.at(ry)) {
      UF.at(rx) = ry;
      SIZE.at(ry) += SIZE.at(rx);
      SIZE.at(rx) = 0;
    }
    else {
      UF.at(ry) = rx;
      SIZE.at(rx) += SIZE.at(ry);
      SIZE.at(ry) = 0;
    }
  }
  bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
    int rx = root(x);
    int ry = root(y);
    return rx == ry;
  }
  int size(int x) { // xと連結されてる要素の数を返す
    return SIZE.at(root(x));
  }
};

//関数 (計算量)
//ctoi (O(1))
int ctoi(char c){
  if (c == '0') return 0;
  if (c == '1') return 1;
  if (c == '2') return 2;
  if (c == '3') return 3;
  if (c == '4') return 4;
  if (c == '5') return 5; 
  if (c == '6') return 6;
  if (c == '7') return 7;
  if (c == '8') return 8;
  if (c == '9') return 9;
  return -1;  
}
//to_char (O(1))
char to_char(int i){
  if (i == 0) return '0';
  if (i == 1) return '1';
  if (i == 2) return '2';
  if (i == 3) return '3';
  if (i == 4) return '4';
  if (i == 5) return '5';
  if (i == 6) return '6';
  if (i == 7) return '7';
  if (i == 8) return '8'; 
  if (i == 9) return '9';
  return ' ';  
}
//二分探索(その要素があるかないか判定するだけならsetとか二分木使う) (0(NlogN))
ll BS(vector<ll> V, ll Q) {
  sort(V.begin(), V.end()); // こことれば(O(logN))
  int L = -1; // Q未満の最大の要素
  int R = V.size(); // Q以上の最小の要素
  while (R - L > 1) {
    int M = (L + R) / 2;
    if (V.at(M) < Q) L = M;
    else R = M;
  }
  if (R == int(V.size())) return INF; // Qより大きい要素がない時はINFを返す
  return V.at(R); // Lを返すとQ未満の最大の要素が出てくる。
}
//素数判定 
//一つ一つの判定ではこっちの方が速い(一つ当たりO(√N))
//一方100までに何個あるかなどはEratosthenesの篩の方が良い（O(NlogN))
bool PN(int x) {
  if (x <= 1) return false; // 1や0,-1は素数ではない
  if (x == 2) return true; // √2 + 1 > 2 なので下でやると 2 ％ 2 = 0 となり判定できない
  for (int i = 2; i < sqrt(x) + 1; i++) {
    if (x % i == 0) return false; // 割れたら素数じゃない
  }
  return true;
}
// A^N mod M を計算する。modしたくなかったらM = LINFとかにすればよい (O(√N))
ll modpow(ll A, ll N, ll M) {
  ll ans = 1;
  while (N > 0) {
    if (N & 1) ans = ans * A % M;
      A = A * A % M;
      N >>= 1;
   }
  return ans;
}
  
//メイン処理
int main() {
  cout << fixed << setprecision(16); //精度向上
  //ここまでテンプレ
  string O, E;
  cin >> O >> E;
  for (int i = 0; i < int(O.size()); i++) {
    cout << O.at(i);
    if (i == int(E.size())) {
      break;
    }
    cout << E.at(i);
  }
  cout << endl;
}