#include <bits/stdc++.h>
using namespace std;
int n, m;
vector<long long> Invert(long long x) {
  vector<long long> ret;
  for (long long j = 0;; j++) {
    long long xt = (j * (j - 1)) / 2;
    if (xt == x) ret.push_back(j);
    if (xt > x) break;
  }
  return ret;
}
string Solve(long long z, long long o, long long a01, long long a10) {
  if (a01 + a10 != z * o) return "";
  vector<int> num_zeros_before;
  for (int j = 0; j < o; j++) {
    num_zeros_before.push_back(z - min(z, a10));
    a10 -= min(z, a10);
  }
  string ans;
  ans.resize(z + o);
  for (int j = 0; j < z + o; j++) ans[j] = '?';
  for (int j = 0; j < num_zeros_before.size(); j++) {
    ans[num_zeros_before[j] + j] = '1';
  }
  for (int j = 0; j < z + o; j++)
    if (ans[j] == '?') ans[j] = '0';
  return ans;
}
int main() {
  while (0) {
    string S;
    cin >> S;
    int num_inv = 0;
    for (int i = 0; i < S.size(); i++)
      for (int j = i + 1; j < S.size(); j++) {
        if (S[i] == '1' && S[j] == '0') num_inv++;
      }
    cout << num_inv << endl;
  }
  long long a00, a01, a10, a11;
  cin >> a00 >> a01 >> a10 >> a11;
  vector<long long> num_zeros = Invert(a00);
  vector<long long> num_ones = Invert(a11);
  string answer;
  for (int i = 0; i < num_zeros.size(); i++) {
    for (int j = 0; j < num_ones.size(); j++) {
      string curr = Solve(num_zeros[i], num_ones[j], a01, a10);
      if (!curr.empty()) {
        answer = curr;
        break;
      }
    }
    if (!answer.empty()) break;
  }
  if (answer.empty()) {
    cout << "Impossible" << endl;
  } else {
    cout << answer << endl;
  }
}

### Prompt

In cpp, your task is to solve the following problem:

For each string s consisting of characters '0' and '1' one can define four integers a00, a01, a10 and a11, where axy is the number of subsequences of length 2 of the string s equal to the sequence {x, y}. 

In these problem you are given four integers a00, a01, a10, a11 and have to find any non-empty string s that matches them, or determine that there is no such string. One can prove that if at least one answer exists, there exists an answer of length no more than 1 000 000.

Input

The only line of the input contains four non-negative integers a00, a01, a10 and a11. Each of them doesn't exceed 109.

Output

If there exists a non-empty string that matches four integers from the input, print it in the only line of the output. Otherwise, print "Impossible". The length of your answer must not exceed 1 000 000.

Examples

Input

1 2 3 4


Output

Impossible


Input

1 2 2 1


Output

0110

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m;
vector<long long> Invert(long long x) {
  vector<long long> ret;
  for (long long j = 0;; j++) {
    long long xt = (j * (j - 1)) / 2;
    if (xt == x) ret.push_back(j);
    if (xt > x) break;
  }
  return ret;
}
string Solve(long long z, long long o, long long a01, long long a10) {
  if (a01 + a10 != z * o) return "";
  vector<int> num_zeros_before;
  for (int j = 0; j < o; j++) {
    num_zeros_before.push_back(z - min(z, a10));
    a10 -= min(z, a10);
  }
  string ans;
  ans.resize(z + o);
  for (int j = 0; j < z + o; j++) ans[j] = '?';
  for (int j = 0; j < num_zeros_before.size(); j++) {
    ans[num_zeros_before[j] + j] = '1';
  }
  for (int j = 0; j < z + o; j++)
    if (ans[j] == '?') ans[j] = '0';
  return ans;
}
int main() {
  while (0) {
    string S;
    cin >> S;
    int num_inv = 0;
    for (int i = 0; i < S.size(); i++)
      for (int j = i + 1; j < S.size(); j++) {
        if (S[i] == '1' && S[j] == '0') num_inv++;
      }
    cout << num_inv << endl;
  }
  long long a00, a01, a10, a11;
  cin >> a00 >> a01 >> a10 >> a11;
  vector<long long> num_zeros = Invert(a00);
  vector<long long> num_ones = Invert(a11);
  string answer;
  for (int i = 0; i < num_zeros.size(); i++) {
    for (int j = 0; j < num_ones.size(); j++) {
      string curr = Solve(num_zeros[i], num_ones[j], a01, a10);
      if (!curr.empty()) {
        answer = curr;
        break;
      }
    }
    if (!answer.empty()) break;
  }
  if (answer.empty()) {
    cout << "Impossible" << endl;
  } else {
    cout << answer << endl;
  }
}

```