#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a0, a1, b0, b1, n, m, s1, s2;
  cin >> a0 >> b0 >> b1 >> a1;
  if (a0 == 0 && b0 == 0 && b1 == 0 && a1 == 0) {
    printf("1\n");
    return 0;
  }
  for (s1 = n = 0; s1 + n <= a0; s1 += n, n++)
    ;
  for (s2 = m = 0; s2 + m <= a1; s2 += m, m++)
    ;
  if (s1 != a0 || s2 != a1) {
    printf("Impossible\n");
    return 0;
  }
  if (a0 == 0 && b0 == 0 && b1 == 0) n = 0;
  if (b0 == 0 && b1 == 0 && a1 == 0) m = 0;
  if (b0 + b1 != n * m) {
    printf("Impossible\n");
    return 0;
  }
  for (; n > 0 && m > 0;) {
    if (b0 >= m) {
      printf("0");
      n--;
      b0 -= m;
    } else {
      printf("1");
      m--;
    }
  }
  while (n--) printf("0");
  while (m--) printf("1");
  printf("\n");
  return 0;
}

### Prompt

Generate a cpp solution to 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 a0, a1, b0, b1, n, m, s1, s2;
  cin >> a0 >> b0 >> b1 >> a1;
  if (a0 == 0 && b0 == 0 && b1 == 0 && a1 == 0) {
    printf("1\n");
    return 0;
  }
  for (s1 = n = 0; s1 + n <= a0; s1 += n, n++)
    ;
  for (s2 = m = 0; s2 + m <= a1; s2 += m, m++)
    ;
  if (s1 != a0 || s2 != a1) {
    printf("Impossible\n");
    return 0;
  }
  if (a0 == 0 && b0 == 0 && b1 == 0) n = 0;
  if (b0 == 0 && b1 == 0 && a1 == 0) m = 0;
  if (b0 + b1 != n * m) {
    printf("Impossible\n");
    return 0;
  }
  for (; n > 0 && m > 0;) {
    if (b0 >= m) {
      printf("0");
      n--;
      b0 -= m;
    } else {
      printf("1");
      m--;
    }
  }
  while (n--) printf("0");
  while (m--) printf("1");
  printf("\n");
  return 0;
}

```