#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T lowbit(T x) {
  return x & (-x);
}
double delta(int x) { return sqrt(1.0 + 8.0 * x); }
int main(void) {
  int a00, a01, a10, a11;
  cin >> a00 >> a01 >> a10 >> a11;
  int n = (1 + delta(a00)) / 2 + 0.5, m = (1 + delta(a11)) / 2 + 0.5;
  bool flag = true;
  if (n * (n - 1) / 2 != a00) flag = false;
  if (m * (m - 1) / 2 != a11) flag = false;
  if (n == 1 && a10 == 0 && a01 == 0)
    n = 0;
  else if (m == 1 && a10 == 0 && a01 == 0)
    m = 0;
  if (a01 + a10 != n * m) flag = false;
  int len = n + m;
  if (flag) {
    for (int i = 0; i < len; i++) {
      if (a01 >= m) {
        printf("0");
        a01 -= m;
        n--;
      } else if (a10 >= n) {
        printf("1");
        a10 -= n;
        m--;
      }
    }
    puts("");
  } else {
    printf("Impossible\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;
template <class T>
inline T lowbit(T x) {
  return x & (-x);
}
double delta(int x) { return sqrt(1.0 + 8.0 * x); }
int main(void) {
  int a00, a01, a10, a11;
  cin >> a00 >> a01 >> a10 >> a11;
  int n = (1 + delta(a00)) / 2 + 0.5, m = (1 + delta(a11)) / 2 + 0.5;
  bool flag = true;
  if (n * (n - 1) / 2 != a00) flag = false;
  if (m * (m - 1) / 2 != a11) flag = false;
  if (n == 1 && a10 == 0 && a01 == 0)
    n = 0;
  else if (m == 1 && a10 == 0 && a01 == 0)
    m = 0;
  if (a01 + a10 != n * m) flag = false;
  int len = n + m;
  if (flag) {
    for (int i = 0; i < len; i++) {
      if (a01 >= m) {
        printf("0");
        a01 -= m;
        n--;
      } else if (a10 >= n) {
        printf("1");
        a10 -= n;
        m--;
      }
    }
    puts("");
  } else {
    printf("Impossible\n");
  }
  return 0;
}

```