#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
char s[N];
long long a, b, c, d;
long long cal(long long x) {
  long long y = sqrt(x * 2);
  if (y * (y + 1) / 2 != x) {
    printf("Impossible\n");
    exit(0);
  }
  return y + 1;
}
long long C(long long x) { return 1LL * x * (x - 1) / 2; }
void solve(long long x, long long y, long long c01, long long c10) {
  if (C(x + y) != c01 + c10 + a + d) return;
  for (int i = y, j = 0; j < x + y; j++) {
    if (c01 >= i)
      c01 -= i, s[j] = '0';
    else
      s[j] = '1', i--;
  }
  if (x == 0 && y == 0)
    printf("0");
  else
    printf("%s", s);
  exit(0);
}
int main() {
  cin >> a >> b >> c >> d;
  long long x = cal(a), y = cal(d);
  if (a == 0 && d == 0) {
    solve(0, 0, b, c);
  }
  if (a == 0) {
    solve(0, y, b, c);
  }
  if (d == 0) {
    solve(x, 0, b, c);
  }
  solve(x, y, b, c);
  printf("Impossible\n");
}

### Prompt

Create a solution in CPP for 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;
const int N = 100005;
char s[N];
long long a, b, c, d;
long long cal(long long x) {
  long long y = sqrt(x * 2);
  if (y * (y + 1) / 2 != x) {
    printf("Impossible\n");
    exit(0);
  }
  return y + 1;
}
long long C(long long x) { return 1LL * x * (x - 1) / 2; }
void solve(long long x, long long y, long long c01, long long c10) {
  if (C(x + y) != c01 + c10 + a + d) return;
  for (int i = y, j = 0; j < x + y; j++) {
    if (c01 >= i)
      c01 -= i, s[j] = '0';
    else
      s[j] = '1', i--;
  }
  if (x == 0 && y == 0)
    printf("0");
  else
    printf("%s", s);
  exit(0);
}
int main() {
  cin >> a >> b >> c >> d;
  long long x = cal(a), y = cal(d);
  if (a == 0 && d == 0) {
    solve(0, 0, b, c);
  }
  if (a == 0) {
    solve(0, y, b, c);
  }
  if (d == 0) {
    solve(x, 0, b, c);
  }
  solve(x, y, b, c);
  printf("Impossible\n");
}

```