#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, t, cnt1, cnt2, num[1000005];
int main() {
  scanf("%d%d%d%d", &a, &b, &c, &d);
  cnt1 = sqrt(2 * a) + 1, cnt2 = sqrt(2 * d) + 1;
  if (cnt1 * (cnt1 - 1) != 2 * a || cnt2 * (cnt2 - 1) != 2 * d)
    return puts("Impossible"), 0;
  if (a + b + c + d == 0) return puts("0"), 0;
  if (a + b + c == 0) {
    for (int i = 1; i <= cnt2; i++) printf("1");
    return 0;
  }
  if (b + c + d == 0) {
    for (int i = 1; i <= cnt1; i++) printf("0");
    return 0;
  }
  if (cnt1 * cnt2 != b + c) return puts("Impossible"), 0;
  while (c) {
    if (c >= cnt1)
      c -= cnt1, num[++t] = 1;
    else {
      num[++t + cnt1 - c] = 1;
      c = 0;
      break;
    }
  }
  for (int i = 1; i <= cnt2 - t; i++) num[cnt1 + cnt2 + 1 - i] = 1;
  for (int i = 1; i <= cnt1 + cnt2; i++) printf("%d", num[i]);
  return 0;
}

### Prompt

Your challenge is to write 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;
int a, b, c, d, t, cnt1, cnt2, num[1000005];
int main() {
  scanf("%d%d%d%d", &a, &b, &c, &d);
  cnt1 = sqrt(2 * a) + 1, cnt2 = sqrt(2 * d) + 1;
  if (cnt1 * (cnt1 - 1) != 2 * a || cnt2 * (cnt2 - 1) != 2 * d)
    return puts("Impossible"), 0;
  if (a + b + c + d == 0) return puts("0"), 0;
  if (a + b + c == 0) {
    for (int i = 1; i <= cnt2; i++) printf("1");
    return 0;
  }
  if (b + c + d == 0) {
    for (int i = 1; i <= cnt1; i++) printf("0");
    return 0;
  }
  if (cnt1 * cnt2 != b + c) return puts("Impossible"), 0;
  while (c) {
    if (c >= cnt1)
      c -= cnt1, num[++t] = 1;
    else {
      num[++t + cnt1 - c] = 1;
      c = 0;
      break;
    }
  }
  for (int i = 1; i <= cnt2 - t; i++) num[cnt1 + cnt2 + 1 - i] = 1;
  for (int i = 1; i <= cnt1 + cnt2; i++) printf("%d", num[i]);
  return 0;
}

```