#include <bits/stdc++.h>
using namespace std;
long long a00, a01, a10, a11;
long long sum0, sum1;
int main() {
  cin >> a00 >> a01 >> a10 >> a11;
  if (a00) {
    sum0 = sqrt(a00 * 2) + 1;
  } else {
    sum0 = 0;
  }
  if (a11) {
    sum1 = sqrt(a11 * 2) + 1;
  } else {
    sum1 = 0;
  }
  if (a00) {
    if (sum0 * (sum0 - 1) / 2 != a00) {
      printf("Impossible\n");
      return 0;
    }
  }
  if (a11) {
    if (sum1 * (sum1 - 1) / 2 != a11) {
      printf("Impossible\n");
      return 0;
    }
  }
  if (a00 == 0 && a11 == 0) {
    if (a10 == 0 && a01 == 0) {
      printf("1\n");
      return 0;
    }
    if (a10 == 1 && a01 == 0) {
      printf("10\n");
      return 0;
    }
    if (a10 == 0 && a01 == 1) {
      printf("01\n");
      return 0;
    }
    printf("Impossible\n");
    return 0;
  }
  if (a00 == 0) {
    if (a01 + a10 == sum1) {
      for (int i = 1; i <= a10; i++) printf("1");
      printf("0");
      for (int i = 1; i <= a01; i++) printf("1");
      printf("\n");
      return 0;
    }
    if (a01 + a10 == 0) {
      for (int i = 1; i <= sum1; i++) printf("1");
      printf("\n");
      return 0;
    }
    printf("Impossible\n");
    return 0;
  }
  if (a11 == 0) {
    if (a01 + a10 == sum0) {
      for (int i = 1; i <= a01; i++) printf("0");
      printf("1");
      for (int i = 1; i <= a10; i++) printf("0");
      printf("\n");
      return 0;
    }
    if (a01 + a10 == 0) {
      for (int i = 1; i <= sum0; i++) printf("0");
      printf("\n");
      return 0;
    }
    printf("Impossible\n");
    return 0;
  }
  if ((sum0 * (sum0 - 1) / 2 == a00) && (sum1 * (sum1 - 1) / 2 == a11) &&
      (sum0 * sum1 == a01 + a10)) {
  } else {
    printf("Impossible\n");
    return 0;
  }
  long long nex0 = sum0, nex1 = sum1;
  for (int i = 1; i <= sum0 + sum1; i++) {
    if (nex1 <= a01) {
      printf("0");
      a01 -= nex1;
      nex0--;
    } else {
      printf("1");
      a10 -= nex0;
      nex1--;
    }
  }
  printf("\n");
  return 0;
}

### Prompt

Generate 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;
long long a00, a01, a10, a11;
long long sum0, sum1;
int main() {
  cin >> a00 >> a01 >> a10 >> a11;
  if (a00) {
    sum0 = sqrt(a00 * 2) + 1;
  } else {
    sum0 = 0;
  }
  if (a11) {
    sum1 = sqrt(a11 * 2) + 1;
  } else {
    sum1 = 0;
  }
  if (a00) {
    if (sum0 * (sum0 - 1) / 2 != a00) {
      printf("Impossible\n");
      return 0;
    }
  }
  if (a11) {
    if (sum1 * (sum1 - 1) / 2 != a11) {
      printf("Impossible\n");
      return 0;
    }
  }
  if (a00 == 0 && a11 == 0) {
    if (a10 == 0 && a01 == 0) {
      printf("1\n");
      return 0;
    }
    if (a10 == 1 && a01 == 0) {
      printf("10\n");
      return 0;
    }
    if (a10 == 0 && a01 == 1) {
      printf("01\n");
      return 0;
    }
    printf("Impossible\n");
    return 0;
  }
  if (a00 == 0) {
    if (a01 + a10 == sum1) {
      for (int i = 1; i <= a10; i++) printf("1");
      printf("0");
      for (int i = 1; i <= a01; i++) printf("1");
      printf("\n");
      return 0;
    }
    if (a01 + a10 == 0) {
      for (int i = 1; i <= sum1; i++) printf("1");
      printf("\n");
      return 0;
    }
    printf("Impossible\n");
    return 0;
  }
  if (a11 == 0) {
    if (a01 + a10 == sum0) {
      for (int i = 1; i <= a01; i++) printf("0");
      printf("1");
      for (int i = 1; i <= a10; i++) printf("0");
      printf("\n");
      return 0;
    }
    if (a01 + a10 == 0) {
      for (int i = 1; i <= sum0; i++) printf("0");
      printf("\n");
      return 0;
    }
    printf("Impossible\n");
    return 0;
  }
  if ((sum0 * (sum0 - 1) / 2 == a00) && (sum1 * (sum1 - 1) / 2 == a11) &&
      (sum0 * sum1 == a01 + a10)) {
  } else {
    printf("Impossible\n");
    return 0;
  }
  long long nex0 = sum0, nex1 = sum1;
  for (int i = 1; i <= sum0 + sum1; i++) {
    if (nex1 <= a01) {
      printf("0");
      a01 -= nex1;
      nex0--;
    } else {
      printf("1");
      a10 -= nex0;
      nex1--;
    }
  }
  printf("\n");
  return 0;
}

```