#include <bits/stdc++.h>
using namespace std;
long long f(long long x) { return x * (x - 1) / 2; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a, b, c, d;
  cin >> a >> b >> c >> d;
  long long n = -1, m = -1;
  for (long long i = 0; f(i) <= a; i++)
    if (f(i) == a) {
      for (long long j = 0; f(j) <= d; j++)
        if (f(j) == d) {
          if (i * j == b + c) {
            n = i, m = j;
            break;
          }
        }
      if (n != -1) break;
    }
  if (n == -1) {
    cout << "Impossible" << endl;
    return 0;
  }
  if (n == 0 && m == 0) {
    cout << 0 << endl;
    return 0;
  }
  long long s = n + m;
  for (long long i = 0; i < s; i++) {
    if (c >= n) {
      m--;
      c -= n;
      cout << 1;
    } else {
      n--;
      cout << 0;
    }
  }
  cout << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long long f(long long x) { return x * (x - 1) / 2; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a, b, c, d;
  cin >> a >> b >> c >> d;
  long long n = -1, m = -1;
  for (long long i = 0; f(i) <= a; i++)
    if (f(i) == a) {
      for (long long j = 0; f(j) <= d; j++)
        if (f(j) == d) {
          if (i * j == b + c) {
            n = i, m = j;
            break;
          }
        }
      if (n != -1) break;
    }
  if (n == -1) {
    cout << "Impossible" << endl;
    return 0;
  }
  if (n == 0 && m == 0) {
    cout << 0 << endl;
    return 0;
  }
  long long s = n + m;
  for (long long i = 0; i < s; i++) {
    if (c >= n) {
      m--;
      c -= n;
      cout << 1;
    } else {
      n--;
      cout << 0;
    }
  }
  cout << endl;
  return 0;
}

```