#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a00, a01, a10, a11;
  scanf("%I64d %I64d %I64d %I64d", &a00, &a01, &a10, &a11);
  if (a00 == 0 && a01 == 0 && a10 == 0 && a11 == 0) {
    printf("1");
    return 0;
  }
  long long c0 = 0;
  while (c0 * (c0 - 1) / 2 < a00) c0++;
  if (c0 * (c0 - 1) / 2 != a00) {
    printf("Impossible\n");
    return 0;
  }
  long long c1 = 0;
  while (c1 * (c1 - 1) / 2 < a11) c1++;
  if (c1 * (c1 - 1) / 2 != a11) {
    printf("Impossible\n");
    return 0;
  }
  if (a10 + a01 > 0) {
    if (c0 == 0) c0 = 1;
    if (c1 == 0) c1 = 1;
  }
  if (c0 * c1 != a10 + a01) {
    printf("Impossible\n");
    return 0;
  }
  if (c0 == 0) {
    for (int i = 0; i < c1; i++) printf("1");
    return 0;
  }
  if (c1 == 0) {
    for (int i = 0; i < c0; i++) printf("0");
    return 0;
  }
  for (int i = 0; i < a01 / c1; i++) printf("0");
  if (a01 % c1 == 0) {
    for (int i = 0; i < c1; i++) printf("1");
    for (int i = 0; i < c0 - a01 / c1; i++) printf("0");
  } else {
    for (int i = 0; i < c1 - a01 % c1; i++) printf("1");
    printf("0");
    for (int i = 0; i < a01 % c1; i++) printf("1");
    for (int i = 0; i < c0 - a01 / c1 - 1; i++) printf("0");
  }
  return 0;
}

### 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;
int main() {
  long long a00, a01, a10, a11;
  scanf("%I64d %I64d %I64d %I64d", &a00, &a01, &a10, &a11);
  if (a00 == 0 && a01 == 0 && a10 == 0 && a11 == 0) {
    printf("1");
    return 0;
  }
  long long c0 = 0;
  while (c0 * (c0 - 1) / 2 < a00) c0++;
  if (c0 * (c0 - 1) / 2 != a00) {
    printf("Impossible\n");
    return 0;
  }
  long long c1 = 0;
  while (c1 * (c1 - 1) / 2 < a11) c1++;
  if (c1 * (c1 - 1) / 2 != a11) {
    printf("Impossible\n");
    return 0;
  }
  if (a10 + a01 > 0) {
    if (c0 == 0) c0 = 1;
    if (c1 == 0) c1 = 1;
  }
  if (c0 * c1 != a10 + a01) {
    printf("Impossible\n");
    return 0;
  }
  if (c0 == 0) {
    for (int i = 0; i < c1; i++) printf("1");
    return 0;
  }
  if (c1 == 0) {
    for (int i = 0; i < c0; i++) printf("0");
    return 0;
  }
  for (int i = 0; i < a01 / c1; i++) printf("0");
  if (a01 % c1 == 0) {
    for (int i = 0; i < c1; i++) printf("1");
    for (int i = 0; i < c0 - a01 / c1; i++) printf("0");
  } else {
    for (int i = 0; i < c1 - a01 % c1; i++) printf("1");
    printf("0");
    for (int i = 0; i < a01 % c1; i++) printf("1");
    for (int i = 0; i < c0 - a01 / c1 - 1; i++) printf("0");
  }
  return 0;
}

```