#include <bits/stdc++.h>
using namespace std;
long long a, b, c, d, n0, n1, siz;
vector<long long> str;
long long fin(long long x) {
  long long ans = -1;
  for (long long i = 0; i <= 2 * x; i++) {
    if (i * (i - 1) == 2 * x) {
      ans = i;
      break;
    }
    if (i * (i - 1) > 2 * x) break;
  }
  return ans;
}
void sol() {
  if (b + c != n0 * n1) {
    cout << "Impossible";
    return;
  }
  while (b) {
    if (b >= n1) {
      cout << 0;
      b -= n1;
      n0--;
      continue;
    }
    if (b < n1) {
      cout << 1;
      n1--;
      continue;
    }
  }
  while (n1--) cout << 1;
  while (n0--) cout << 0;
}
int main() {
  cin >> a >> b >> c >> d;
  n0 = fin(a);
  n1 = fin(d);
  siz = n1 + n0;
  if (n0 == -1 || n1 == -1) {
    cout << "Impossible";
    return 0;
  }
  if (n0 == 0 && n1 == 0 && (b + c) == 0) {
    cout << 0;
    return 0;
  }
  if (n0 == 0 && (b + c) > 0) n0 = 1;
  if (n1 == 0 && (b + c) > 0) n1 = 1;
  sol();
}

### Prompt

Construct a cpp code solution to the problem outlined:

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, n0, n1, siz;
vector<long long> str;
long long fin(long long x) {
  long long ans = -1;
  for (long long i = 0; i <= 2 * x; i++) {
    if (i * (i - 1) == 2 * x) {
      ans = i;
      break;
    }
    if (i * (i - 1) > 2 * x) break;
  }
  return ans;
}
void sol() {
  if (b + c != n0 * n1) {
    cout << "Impossible";
    return;
  }
  while (b) {
    if (b >= n1) {
      cout << 0;
      b -= n1;
      n0--;
      continue;
    }
    if (b < n1) {
      cout << 1;
      n1--;
      continue;
    }
  }
  while (n1--) cout << 1;
  while (n0--) cout << 0;
}
int main() {
  cin >> a >> b >> c >> d;
  n0 = fin(a);
  n1 = fin(d);
  siz = n1 + n0;
  if (n0 == -1 || n1 == -1) {
    cout << "Impossible";
    return 0;
  }
  if (n0 == 0 && n1 == 0 && (b + c) == 0) {
    cout << 0;
    return 0;
  }
  if (n0 == 0 && (b + c) > 0) n0 = 1;
  if (n1 == 0 && (b + c) > 0) n1 = 1;
  sol();
}

```