#include <bits/stdc++.h>
using namespace std;
int a00, a01, a10, a11;
int n0, n1;
int main() {
  scanf("%d%d%d%d", &a00, &a01, &a10, &a11);
  int b1 = 0, b2 = 0;
  n0 = 1;
  n1 = 1;
  if (a00 == 0 && a01 == 0 && a10 == 0 && a11 == 0) {
    printf("0\n");
    return 0;
  }
  while (b1 < a00) {
    b1 += n0;
    n0++;
  }
  while (b2 < a11) {
    b2 += n1;
    n1++;
  }
  if (a00 == 0 && a01 == 0 && a10 == 0) n0 = 0;
  if (a11 == 0 && a01 == 0 && a10 == 0) n1 = 0;
  if (b1 != a00 || b2 != a11 || (a01 + a10 != n1 * n0)) {
    printf("Impossible\n");
    return 0;
  }
  if (a00 == 0 && a10 == 0 && a01 == 0) {
    for (int i = 0; i < n1; i++) printf("1");
    printf("\n");
    return 0;
  }
  if (a11 == 0 && a10 == 0 && a01 == 0) {
    for (int i = 0; i < n0; i++) printf("0");
    printf("\n");
    return 0;
  }
  int num1 = a01 / n1;
  int num2 = a01 % n1;
  for (int i = 0; i < num1; i++) printf("0");
  for (int i = 0; i < n1 - num2; i++) printf("1");
  if (num1 != n0) printf("0");
  for (int i = 0; i < num2; i++) printf("1");
  for (int i = 0; i < n0 - num1 - 1; i++) printf("0");
  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;
int a00, a01, a10, a11;
int n0, n1;
int main() {
  scanf("%d%d%d%d", &a00, &a01, &a10, &a11);
  int b1 = 0, b2 = 0;
  n0 = 1;
  n1 = 1;
  if (a00 == 0 && a01 == 0 && a10 == 0 && a11 == 0) {
    printf("0\n");
    return 0;
  }
  while (b1 < a00) {
    b1 += n0;
    n0++;
  }
  while (b2 < a11) {
    b2 += n1;
    n1++;
  }
  if (a00 == 0 && a01 == 0 && a10 == 0) n0 = 0;
  if (a11 == 0 && a01 == 0 && a10 == 0) n1 = 0;
  if (b1 != a00 || b2 != a11 || (a01 + a10 != n1 * n0)) {
    printf("Impossible\n");
    return 0;
  }
  if (a00 == 0 && a10 == 0 && a01 == 0) {
    for (int i = 0; i < n1; i++) printf("1");
    printf("\n");
    return 0;
  }
  if (a11 == 0 && a10 == 0 && a01 == 0) {
    for (int i = 0; i < n0; i++) printf("0");
    printf("\n");
    return 0;
  }
  int num1 = a01 / n1;
  int num2 = a01 % n1;
  for (int i = 0; i < num1; i++) printf("0");
  for (int i = 0; i < n1 - num2; i++) printf("1");
  if (num1 != n0) printf("0");
  for (int i = 0; i < num2; i++) printf("1");
  for (int i = 0; i < n0 - num1 - 1; i++) printf("0");
  printf("\n");
  return 0;
}

```