#include <bits/stdc++.h>
using namespace std;
long long a00, a01, a10, a11;
int main() {
  scanf("%lld%lld%lld%lld", &a00, &a01, &a10, &a11);
  long long n = 1;
  while (n <= 1000000) {
    if (a00 + a01 + a10 + a11 == n * (n - 1) / 2)
      break;
    else
      n++;
  }
  if (n > 1000000) {
    printf("Impossible");
    return 0;
  }
  long long zeros = -1, ones = -1;
  for (long long i = 1; i <= n; i++) {
    if (i * (i - 1) / 2 == a00) zeros = i;
    if (i * (i - 1) / 2 == a11) ones = i;
  }
  if (zeros + ones > n) {
    if (zeros == 1)
      zeros--;
    else if (ones == 1)
      ones--;
  }
  if (zeros + ones != n) {
    printf("Impossible");
    return 0;
  }
  if (zeros < 0 || ones < 0) {
    printf("Impossible");
    return 0;
  }
  if (zeros == 0) {
    for (int i = 0; i < ones; i++) printf("1");
    return 0;
  }
  if (ones == 0) {
    for (int i = 0; i < zeros; i++) printf("0");
    return 0;
  }
  long long z = a01 / zeros;
  long long m = a01 % zeros;
  long long lft = m ? ones - z - 1 : ones - z;
  for (int i = 0; i < lft; i++) printf("1");
  for (int i = 0; i < m; i++) printf("0");
  if (m) printf("1");
  for (int i = m; i < zeros; i++) printf("0");
  for (int i = 0; i < z; i++) printf("1");
}

### Prompt

Your task is to create 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;
int main() {
  scanf("%lld%lld%lld%lld", &a00, &a01, &a10, &a11);
  long long n = 1;
  while (n <= 1000000) {
    if (a00 + a01 + a10 + a11 == n * (n - 1) / 2)
      break;
    else
      n++;
  }
  if (n > 1000000) {
    printf("Impossible");
    return 0;
  }
  long long zeros = -1, ones = -1;
  for (long long i = 1; i <= n; i++) {
    if (i * (i - 1) / 2 == a00) zeros = i;
    if (i * (i - 1) / 2 == a11) ones = i;
  }
  if (zeros + ones > n) {
    if (zeros == 1)
      zeros--;
    else if (ones == 1)
      ones--;
  }
  if (zeros + ones != n) {
    printf("Impossible");
    return 0;
  }
  if (zeros < 0 || ones < 0) {
    printf("Impossible");
    return 0;
  }
  if (zeros == 0) {
    for (int i = 0; i < ones; i++) printf("1");
    return 0;
  }
  if (ones == 0) {
    for (int i = 0; i < zeros; i++) printf("0");
    return 0;
  }
  long long z = a01 / zeros;
  long long m = a01 % zeros;
  long long lft = m ? ones - z - 1 : ones - z;
  for (int i = 0; i < lft; i++) printf("1");
  for (int i = 0; i < m; i++) printf("0");
  if (m) printf("1");
  for (int i = m; i < zeros; i++) printf("0");
  for (int i = 0; i < z; i++) printf("1");
}

```