#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a00, a01, a10, a11;
  scanf("%I64d%I64d%I64d%I64d", &a00, &a01, &a10, &a11);
  long long g0, g1;
  g0 = (1 + sqrt(1 + 8 * a00)) / 2;
  g1 = (1 + sqrt(1 + 8 * a11)) / 2;
  if (a00 + a01 + a10 + a11 == 0) {
    printf("0\n");
    return 0;
  }
  if (g0 == 1 || g1 == 1) {
    if (g0 == 1) {
      if (a01 || a10)
        g0 = 1;
      else
        g0 = 0;
    }
    if (g1 == 1) {
      if (a01 || a10)
        g1 = 1;
      else
        g1 = 0;
    }
  }
  if (g0 * (g0 - 1) != 2 * a00 || g1 * (g1 - 1) != 2 * a11) {
    printf("Impossible\n");
    return 0;
  }
  long long gz = g0 + g1;
  if (gz * (gz - 1) / 2 != (a00 + a01 + a10 + a11)) {
    printf("Impossible\n");
    return 0;
  }
  while (g0 + g1) {
    if (a10 >= g0) {
      printf("1");
      a10 -= g0;
      g1--;
    } else {
      printf("0");
      a01 -= g1;
      g0--;
    }
  }
  printf("\n");
  return 0;
}

### Prompt

Create a solution in cpp for 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 main() {
  long long a00, a01, a10, a11;
  scanf("%I64d%I64d%I64d%I64d", &a00, &a01, &a10, &a11);
  long long g0, g1;
  g0 = (1 + sqrt(1 + 8 * a00)) / 2;
  g1 = (1 + sqrt(1 + 8 * a11)) / 2;
  if (a00 + a01 + a10 + a11 == 0) {
    printf("0\n");
    return 0;
  }
  if (g0 == 1 || g1 == 1) {
    if (g0 == 1) {
      if (a01 || a10)
        g0 = 1;
      else
        g0 = 0;
    }
    if (g1 == 1) {
      if (a01 || a10)
        g1 = 1;
      else
        g1 = 0;
    }
  }
  if (g0 * (g0 - 1) != 2 * a00 || g1 * (g1 - 1) != 2 * a11) {
    printf("Impossible\n");
    return 0;
  }
  long long gz = g0 + g1;
  if (gz * (gz - 1) / 2 != (a00 + a01 + a10 + a11)) {
    printf("Impossible\n");
    return 0;
  }
  while (g0 + g1) {
    if (a10 >= g0) {
      printf("1");
      a10 -= g0;
      g1--;
    } else {
      printf("0");
      a01 -= g1;
      g0--;
    }
  }
  printf("\n");
  return 0;
}

```