#include <bits/stdc++.h>
using namespace std;
int main() {
  int a00, a01, a10, a11;
  while (~scanf("%d%d%d%d", &a00, &a01, &a10, &a11)) {
    if (a00 + a01 + a10 + a11 == 0) {
      cout << "1" << endl;
      continue;
    }
    int l = (int)sqrt(2 * a00) + 1;
    int r = (int)sqrt(2 * a11) + 1;
    if (a01 + a10 == 0) {
      l = a00 ? l : 0;
      r = a11 ? r : 0;
      if (a00 + a11 == 0) l = 1;
    }
    if (l * (l - 1) != 2 * a00 || r * (r - 1) != 2 * a11 ||
        a01 + a10 != l * r) {
      printf("Impossible\n");
      continue;
    }
    while (r > a01) cout << "1", --r;
    while (a01) {
      if (r > a01)
        cout << "1", --r;
      else
        cout << "0", --l, a01 -= r;
    }
    while (r--) cout << "1";
    while (l--) cout << "0";
    printf("\n");
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
int main() {
  int a00, a01, a10, a11;
  while (~scanf("%d%d%d%d", &a00, &a01, &a10, &a11)) {
    if (a00 + a01 + a10 + a11 == 0) {
      cout << "1" << endl;
      continue;
    }
    int l = (int)sqrt(2 * a00) + 1;
    int r = (int)sqrt(2 * a11) + 1;
    if (a01 + a10 == 0) {
      l = a00 ? l : 0;
      r = a11 ? r : 0;
      if (a00 + a11 == 0) l = 1;
    }
    if (l * (l - 1) != 2 * a00 || r * (r - 1) != 2 * a11 ||
        a01 + a10 != l * r) {
      printf("Impossible\n");
      continue;
    }
    while (r > a01) cout << "1", --r;
    while (a01) {
      if (r > a01)
        cout << "1", --r;
      else
        cout << "0", --l, a01 -= r;
    }
    while (r--) cout << "1";
    while (l--) cout << "0";
    printf("\n");
  }
  return 0;
}

```