#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
int main() {
  long long a, b, c, d, zero = 0, one = 0, pp;
  cin >> a >> b >> c >> d;
  if (!a && !b && !c && !d) {
    printf("0\n");
  } else {
    if (b | c) {
      zero = 1;
      one = 1;
    }
    while (zero * (zero - 1) / 2 < a) zero++;
    while (one * (one - 1) / 2 < d) one++;
    if (zero * (zero - 1) / 2 != a) {
      puts("Impossible");
      return 0;
    }
    if (one * (one - 1) / 2 != d) {
      puts("Impossible");
      return 0;
    }
    pp = zero * one;
    if (pp != (b + c)) {
      puts("Impossible");
      return 0;
    }
    pp = zero + one;
    for (int i = 0; i < pp; i++) {
      if (b - one >= 0) {
        b -= one;
        zero--;
        printf("0");
      } else {
        c -= zero;
        one--;
        printf("1");
      }
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = 100000 + 10;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
int main() {
  long long a, b, c, d, zero = 0, one = 0, pp;
  cin >> a >> b >> c >> d;
  if (!a && !b && !c && !d) {
    printf("0\n");
  } else {
    if (b | c) {
      zero = 1;
      one = 1;
    }
    while (zero * (zero - 1) / 2 < a) zero++;
    while (one * (one - 1) / 2 < d) one++;
    if (zero * (zero - 1) / 2 != a) {
      puts("Impossible");
      return 0;
    }
    if (one * (one - 1) / 2 != d) {
      puts("Impossible");
      return 0;
    }
    pp = zero * one;
    if (pp != (b + c)) {
      puts("Impossible");
      return 0;
    }
    pp = zero + one;
    for (int i = 0; i < pp; i++) {
      if (b - one >= 0) {
        b -= one;
        zero--;
        printf("0");
      } else {
        c -= zero;
        one--;
        printf("1");
      }
    }
  }
}

```