#include <bits/stdc++.h>
using namespace std;
long long A[2][2];
long long C[2];
char out[2000000];
int main() {
  for (int i = 0; i < (2); ++i)
    for (int j = 0; j < (2); ++j) cin >> A[i][j];
  for (int i = 0; i < (2); ++i)
    while (C[i] * (C[i] - 1) / 2 < A[i][i]) ++C[i];
  bool ok = true;
  for (int i = 0; i < (2); ++i)
    if (C[i] * (C[i] - 1) / 2 != A[i][i]) ok = false;
  if (A[0][1] || A[1][0]) {
    C[0] = max(C[0], 1LL);
    C[1] = max(C[1], 1LL);
  }
  if (C[0] == 0 && C[1] == 0) C[0] = 1;
  if (A[0][1] + A[1][0] != C[0] * C[1]) ok = false;
  if (!ok) {
    printf("Impossible\n");
    return 0;
  }
  int c = 0;
  while (C[0] || C[1]) {
    if (!C[0] || A[0][1] < C[1]) {
      --C[1];
      out[c++] = '1';
      continue;
    } else {
      --C[0];
      A[0][1] -= C[1];
      out[c++] = '0';
    }
  }
  printf("%s\n", out);
}

### 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;
long long A[2][2];
long long C[2];
char out[2000000];
int main() {
  for (int i = 0; i < (2); ++i)
    for (int j = 0; j < (2); ++j) cin >> A[i][j];
  for (int i = 0; i < (2); ++i)
    while (C[i] * (C[i] - 1) / 2 < A[i][i]) ++C[i];
  bool ok = true;
  for (int i = 0; i < (2); ++i)
    if (C[i] * (C[i] - 1) / 2 != A[i][i]) ok = false;
  if (A[0][1] || A[1][0]) {
    C[0] = max(C[0], 1LL);
    C[1] = max(C[1], 1LL);
  }
  if (C[0] == 0 && C[1] == 0) C[0] = 1;
  if (A[0][1] + A[1][0] != C[0] * C[1]) ok = false;
  if (!ok) {
    printf("Impossible\n");
    return 0;
  }
  int c = 0;
  while (C[0] || C[1]) {
    if (!C[0] || A[0][1] < C[1]) {
      --C[1];
      out[c++] = '1';
      continue;
    } else {
      --C[0];
      A[0][1] -= C[1];
      out[c++] = '0';
    }
  }
  printf("%s\n", out);
}

```