#include <bits/stdc++.h>
using namespace std;
long long a[4], num1, num0;
int main() {
  cin >> a[0] >> a[1] >> a[2] >> a[3];
  if (a[0] == 0 && a[1] == 0 && a[2] == 0 && a[3] == 0) {
    puts("0");
    return 0;
  }
  long long tmp = sqrt(1 + 8 * a[0]);
  if (tmp * tmp != 1 + 8 * a[0] || (1 + tmp) & 1) {
    puts("Impossible");
    return 0;
  }
  num0 = (1 + tmp) >> 1;
  tmp = sqrt(1 + 8 * a[3]);
  if (tmp * tmp != 1 + 8 * a[3] || (1 + tmp) & 1) {
    puts("Impossible");
    return 0;
  }
  num1 = (1 + tmp) >> 1;
  if (a[0] == 0 && a[1] == 0 && a[2] == 0) {
    for (int i = 1; i <= num1; i++) printf("1");
    puts("");
    return 0;
  }
  if (a[1] == 0 && a[2] == 0 && a[3] == 0) {
    for (int i = 1; i <= num0; i++) printf("0");
    puts("");
    return 0;
  }
  if ((long long)num0 * num1 != a[1] + a[2]) {
    puts("Impossible");
    return 0;
  }
  if (num0 == 0) {
    for (int i = 1; i <= num1; i++) printf("1");
    puts("");
    return 0;
  }
  if (num1 == 0) {
    for (int i = 1; i <= num0; i++) printf("0");
    puts("");
    return 0;
  }
  int sum = a[1] / num1, left = a[1] % num1;
  for (int i = 1; i <= sum; i++) printf("0");
  for (int i = 1; i <= num1 - left; i++) printf("1");
  if (num0 - sum) printf("0");
  for (int i = 1; i <= left; i++) printf("1");
  for (int i = sum + 2; i <= num0; i++) printf("0");
  puts("");
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 a[4], num1, num0;
int main() {
  cin >> a[0] >> a[1] >> a[2] >> a[3];
  if (a[0] == 0 && a[1] == 0 && a[2] == 0 && a[3] == 0) {
    puts("0");
    return 0;
  }
  long long tmp = sqrt(1 + 8 * a[0]);
  if (tmp * tmp != 1 + 8 * a[0] || (1 + tmp) & 1) {
    puts("Impossible");
    return 0;
  }
  num0 = (1 + tmp) >> 1;
  tmp = sqrt(1 + 8 * a[3]);
  if (tmp * tmp != 1 + 8 * a[3] || (1 + tmp) & 1) {
    puts("Impossible");
    return 0;
  }
  num1 = (1 + tmp) >> 1;
  if (a[0] == 0 && a[1] == 0 && a[2] == 0) {
    for (int i = 1; i <= num1; i++) printf("1");
    puts("");
    return 0;
  }
  if (a[1] == 0 && a[2] == 0 && a[3] == 0) {
    for (int i = 1; i <= num0; i++) printf("0");
    puts("");
    return 0;
  }
  if ((long long)num0 * num1 != a[1] + a[2]) {
    puts("Impossible");
    return 0;
  }
  if (num0 == 0) {
    for (int i = 1; i <= num1; i++) printf("1");
    puts("");
    return 0;
  }
  if (num1 == 0) {
    for (int i = 1; i <= num0; i++) printf("0");
    puts("");
    return 0;
  }
  int sum = a[1] / num1, left = a[1] % num1;
  for (int i = 1; i <= sum; i++) printf("0");
  for (int i = 1; i <= num1 - left; i++) printf("1");
  if (num0 - sum) printf("0");
  for (int i = 1; i <= left; i++) printf("1");
  for (int i = sum + 2; i <= num0; i++) printf("0");
  puts("");
  return 0;
}

```