#include <bits/stdc++.h>
using namespace std;
int a[4], o, z;
int getNum(int a) {
  int ret = 0;
  while (ret * (ret - 1ll) / 2 < a) ++ret;
  if (ret * (ret - 1ll) / 2 == a) return ret;
  return -1;
}
int main() {
  for (int i = (0); i <= (int)(3); ++i) scanf("%d", &a[i]);
  o = getNum(a[0]), z = getNum(a[3]);
  if (o == -1 || z == -1) return printf("Impossible\n"), 0;
  if (a[1] || a[2]) o = max(o, 1), z = max(z, 1);
  if (o * z != a[1] + a[2]) return printf("Impossible\n"), 0;
  if (!o && !z) printf("0");
  while (o && z) {
    if (a[1] >= z)
      printf("0"), --o, a[1] -= z;
    else if (a[2] >= o)
      printf("1"), --z, a[2] -= o;
  }
  while (o) printf("0"), --o;
  while (z) printf("1"), --z;
  printf("\n");
}

### Prompt

Create a solution in cpp for 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;
int a[4], o, z;
int getNum(int a) {
  int ret = 0;
  while (ret * (ret - 1ll) / 2 < a) ++ret;
  if (ret * (ret - 1ll) / 2 == a) return ret;
  return -1;
}
int main() {
  for (int i = (0); i <= (int)(3); ++i) scanf("%d", &a[i]);
  o = getNum(a[0]), z = getNum(a[3]);
  if (o == -1 || z == -1) return printf("Impossible\n"), 0;
  if (a[1] || a[2]) o = max(o, 1), z = max(z, 1);
  if (o * z != a[1] + a[2]) return printf("Impossible\n"), 0;
  if (!o && !z) printf("0");
  while (o && z) {
    if (a[1] >= z)
      printf("0"), --o, a[1] -= z;
    else if (a[2] >= o)
      printf("1"), --z, a[2] -= o;
  }
  while (o) printf("0"), --o;
  while (z) printf("1"), --z;
  printf("\n");
}

```