#include <bits/stdc++.h>
using namespace std;
int a00, a01, a10, a11;
int num0 = -1, num1 = -1;
int main() {
  scanf("%d%d%d%d", &a00, &a01, &a10, &a11);
  for (int i = 1; i <= 100000; i++) {
    if ((long long)i * (i - 1) / 2 == a00) num0 = i;
    if ((long long)i * (i - 1) / 2 == a11) num1 = i;
  }
  if (num0 == -1 || num1 == -1 || (a01 + a10 != num0 * num1)) {
    if (num0 == 1 && num1 != -1 && (a01 + a10 == 0)) {
      num0 = 0;
    } else if (num0 != -1 && num1 == 1 && (a01 + a10 == 0)) {
      num1 = 0;
    } else {
      printf("Impossible\n");
      return 0;
    }
  }
  while (num0 && a10 >= num0) {
    printf("1");
    num1--;
    a10 -= num0;
  }
  while (num0 > a10) {
    printf("0");
    num0--;
  }
  if (num1) {
    printf("1");
    num1--;
  }
  while (num0) {
    printf("0");
    num0--;
  }
  while (num1) {
    printf("1");
    num1--;
  }
  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>
using namespace std;
int a00, a01, a10, a11;
int num0 = -1, num1 = -1;
int main() {
  scanf("%d%d%d%d", &a00, &a01, &a10, &a11);
  for (int i = 1; i <= 100000; i++) {
    if ((long long)i * (i - 1) / 2 == a00) num0 = i;
    if ((long long)i * (i - 1) / 2 == a11) num1 = i;
  }
  if (num0 == -1 || num1 == -1 || (a01 + a10 != num0 * num1)) {
    if (num0 == 1 && num1 != -1 && (a01 + a10 == 0)) {
      num0 = 0;
    } else if (num0 != -1 && num1 == 1 && (a01 + a10 == 0)) {
      num1 = 0;
    } else {
      printf("Impossible\n");
      return 0;
    }
  }
  while (num0 && a10 >= num0) {
    printf("1");
    num1--;
    a10 -= num0;
  }
  while (num0 > a10) {
    printf("0");
    num0--;
  }
  if (num1) {
    printf("1");
    num1--;
  }
  while (num0) {
    printf("0");
    num0--;
  }
  while (num1) {
    printf("1");
    num1--;
  }
  return 0;
}

```