#include <bits/stdc++.h>
const int N = 1000000 + 10;
void fail() {
  puts("Impossible");
  exit(0);
}
std::map<long long, int> pool;
int a[2][2];
int main() {
  for (int i = 1; i <= 100000; ++i) pool[i * (i - 1LL) / 2] = i;
  scanf("%d%d%d%d", &a[0][0], &a[0][1], &a[1][0], &a[1][1]);
  if (!pool.count(a[0][0])) fail();
  if (!pool.count(a[1][1])) fail();
  long long tot = (long long)a[0][0] + a[0][1] + a[1][0] + a[1][1];
  if (!pool.count(tot)) fail();
  int n = pool[a[0][0]], m = pool[a[1][1]];
  if ((n + m) * (n + m - 1LL) / 2 != tot) {
    if (a[0][0] && a[1][1]) fail();
    if (!a[0][0])
      --n;
    else
      --m;
    if ((n + m) * (n + m - 1LL) / 2 != tot) fail();
  }
  static char ans[N];
  for (int i = 1, t = a[0][1], p = n, q = m; i <= n + m; ++i) {
    if (t >= q && p) {
      --p;
      t -= q;
      ans[i] = '0';
    } else {
      --q;
      ans[i] = '1';
    }
  }
  puts(ans + 1);
  return 0;
}

### Prompt

Please formulate 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>
const int N = 1000000 + 10;
void fail() {
  puts("Impossible");
  exit(0);
}
std::map<long long, int> pool;
int a[2][2];
int main() {
  for (int i = 1; i <= 100000; ++i) pool[i * (i - 1LL) / 2] = i;
  scanf("%d%d%d%d", &a[0][0], &a[0][1], &a[1][0], &a[1][1]);
  if (!pool.count(a[0][0])) fail();
  if (!pool.count(a[1][1])) fail();
  long long tot = (long long)a[0][0] + a[0][1] + a[1][0] + a[1][1];
  if (!pool.count(tot)) fail();
  int n = pool[a[0][0]], m = pool[a[1][1]];
  if ((n + m) * (n + m - 1LL) / 2 != tot) {
    if (a[0][0] && a[1][1]) fail();
    if (!a[0][0])
      --n;
    else
      --m;
    if ((n + m) * (n + m - 1LL) / 2 != tot) fail();
  }
  static char ans[N];
  for (int i = 1, t = a[0][1], p = n, q = m; i <= n + m; ++i) {
    if (t >= q && p) {
      --p;
      t -= q;
      ans[i] = '0';
    } else {
      --q;
      ans[i] = '1';
    }
  }
  puts(ans + 1);
  return 0;
}

```