#include <bits/stdc++.h>
using namespace std;
long long const N = 1e6 + 20;
long long n, a[2][2], t[2], res[N], spos;
inline void get(long long &a, long long b) {
  long long l = 0, r = 1e5;
  while (l < r - 1) {
    long long mid = (l + r) >> 1;
    if (mid * (mid - 1) / 2 <= b)
      l = mid;
    else
      r = mid;
  }
  if (l * (l - 1) / 2 != b) {
    cout << "Impossible\n";
    exit(0);
  }
  a = l;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < 2; i++)
    for (long long j = 0; j < 2; j++) cin >> a[i][j];
  if (!a[0][0] && !a[0][1] && !a[1][0] && !a[1][1]) return cout << "0\n", 0;
  get(t[0], a[0][0]);
  get(t[1], a[1][1]);
  get(n, a[0][0] + a[0][1] + a[1][0] + a[1][1]);
  spos = a[0][1] + a[1][1];
  if (t[0] + t[1] > n) {
    if (!a[0][0] && !a[0][1] && !a[1][0]) t[0]--;
    if (!a[1][1] && !a[0][1] && !a[1][0]) t[1]--;
  }
  if (t[0] + t[1] != n) return cout << "Impossible\n", 0;
  for (long long i = n - 1; i >= 0; i--)
    if (spos >= i && spos - i >= (t[1] - 1) * (t[1] - 2) / 2 && t[1]) {
      spos -= i;
      res[i] = 1;
      t[1]--;
    }
  for (long long i = 0; i < n; i++) cout << res[i];
  cout << '\n';
}

### Prompt

Develop a solution in CPP 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 const N = 1e6 + 20;
long long n, a[2][2], t[2], res[N], spos;
inline void get(long long &a, long long b) {
  long long l = 0, r = 1e5;
  while (l < r - 1) {
    long long mid = (l + r) >> 1;
    if (mid * (mid - 1) / 2 <= b)
      l = mid;
    else
      r = mid;
  }
  if (l * (l - 1) / 2 != b) {
    cout << "Impossible\n";
    exit(0);
  }
  a = l;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < 2; i++)
    for (long long j = 0; j < 2; j++) cin >> a[i][j];
  if (!a[0][0] && !a[0][1] && !a[1][0] && !a[1][1]) return cout << "0\n", 0;
  get(t[0], a[0][0]);
  get(t[1], a[1][1]);
  get(n, a[0][0] + a[0][1] + a[1][0] + a[1][1]);
  spos = a[0][1] + a[1][1];
  if (t[0] + t[1] > n) {
    if (!a[0][0] && !a[0][1] && !a[1][0]) t[0]--;
    if (!a[1][1] && !a[0][1] && !a[1][0]) t[1]--;
  }
  if (t[0] + t[1] != n) return cout << "Impossible\n", 0;
  for (long long i = n - 1; i >= 0; i--)
    if (spos >= i && spos - i >= (t[1] - 1) * (t[1] - 2) / 2 && t[1]) {
      spos -= i;
      res[i] = 1;
      t[1]--;
    }
  for (long long i = 0; i < n; i++) cout << res[i];
  cout << '\n';
}

```