#include <bits/stdc++.h>
using namespace std;
long long a, b, c, d;
long long z, y;
long long ans0, ans1;
int main() {
  cin >> a >> b >> c >> d;
  if (b == 0 && c == 0) {
    if (a == 0 && d == 0) {
      ans0 = 1;
    } else {
      if (a != 0) {
        a = a * 2;
        z = sqrt(a);
        y = a / z;
        if (z * y != a) {
          cout << "Impossible";
          return 0;
        }
        if (z + 1 != y) {
          cout << "Impossible";
          return 0;
        }
        ans0 = y;
      }
      if (d != 0) {
        d = d * 2;
        z = sqrt(d);
        if (z == 0) {
          cout << "Impossible";
          return 0;
        }
        y = d / z;
        if (z * y != d) {
          cout << "Impossible";
          return 0;
        }
        if (z + 1 != y) {
          cout << "Impossible";
          return 0;
        }
        ans1 = y;
      }
    }
  } else {
    a = a * 2;
    z = sqrt(a);
    if (z != 0)
      y = a / z;
    else
      y = 1;
    if (z * y != a) {
      cout << "Impossible";
      return 0;
    }
    if (z + 1 != y) {
      cout << "Impossible";
      return 0;
    }
    ans0 = y;
    d = d * 2;
    z = sqrt(d);
    if (z != 0)
      y = d / z;
    else
      y = 1;
    if (z * y != d) {
      cout << "Impossible";
      return 0;
    }
    if (z + 1 != y) {
      cout << "Impossible";
      return 0;
    }
    ans1 = y;
  }
  z = b + c;
  y = ans0 * ans1;
  if (z != y) {
    cout << "Impossible";
    return 0;
  }
  long long l;
  l = ans1;
  for (int i = 1; i <= ans0; i++) {
    while (l > b) cout << "1", l--;
    cout << "0";
    b -= l;
  }
  while (l != 0) cout << "1", l--;
  return 0;
}

### 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;
long long a, b, c, d;
long long z, y;
long long ans0, ans1;
int main() {
  cin >> a >> b >> c >> d;
  if (b == 0 && c == 0) {
    if (a == 0 && d == 0) {
      ans0 = 1;
    } else {
      if (a != 0) {
        a = a * 2;
        z = sqrt(a);
        y = a / z;
        if (z * y != a) {
          cout << "Impossible";
          return 0;
        }
        if (z + 1 != y) {
          cout << "Impossible";
          return 0;
        }
        ans0 = y;
      }
      if (d != 0) {
        d = d * 2;
        z = sqrt(d);
        if (z == 0) {
          cout << "Impossible";
          return 0;
        }
        y = d / z;
        if (z * y != d) {
          cout << "Impossible";
          return 0;
        }
        if (z + 1 != y) {
          cout << "Impossible";
          return 0;
        }
        ans1 = y;
      }
    }
  } else {
    a = a * 2;
    z = sqrt(a);
    if (z != 0)
      y = a / z;
    else
      y = 1;
    if (z * y != a) {
      cout << "Impossible";
      return 0;
    }
    if (z + 1 != y) {
      cout << "Impossible";
      return 0;
    }
    ans0 = y;
    d = d * 2;
    z = sqrt(d);
    if (z != 0)
      y = d / z;
    else
      y = 1;
    if (z * y != d) {
      cout << "Impossible";
      return 0;
    }
    if (z + 1 != y) {
      cout << "Impossible";
      return 0;
    }
    ans1 = y;
  }
  z = b + c;
  y = ans0 * ans1;
  if (z != y) {
    cout << "Impossible";
    return 0;
  }
  long long l;
  l = ans1;
  for (int i = 1; i <= ans0; i++) {
    while (l > b) cout << "1", l--;
    cout << "0";
    b -= l;
  }
  while (l != 0) cout << "1", l--;
  return 0;
}

```