#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long fh = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= fh;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
long long s[1000005];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a00, a01, a10, a11;
  cin >> a00 >> a01 >> a10 >> a11;
  if (!a00 && !a01 && !a10 && !a11) {
    cout << 0 << endl;
    return 0;
  }
  if ((long long)(sqrt(1 + 8 * a00)) * (long long)(sqrt(1 + 8 * a00)) !=
      1 + 8 * a00) {
    cout << "Impossible" << endl;
    return 0;
  }
  if ((long long)(sqrt(1 + 8 * a11)) * (long long)(sqrt(1 + 8 * a11)) !=
      1 + 8 * a11) {
    cout << "Impossible" << endl;
    return 0;
  }
  long long n0 = (1 + sqrt(1 + 8 * a00)) / 2, n1 = (1 + sqrt(1 + 8 * a11)) / 2;
  if (a00 == 0) {
    if (!a01 && !a10) {
      for (long long i = 1; i <= n1; i++) cout << 1;
      cout << endl;
      return 0;
    }
  }
  if (a11 == 0) {
    if (!a01 && !a10) {
      for (long long i = 1; i <= n0; i++) cout << 0;
      cout << endl;
      return 0;
    }
  }
  if (n0 * n1 != a01 + a10) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (long long i = 1; i <= n0 + n1; i++) s[i] = 1;
  long long cyc = a10 / n1, rem = a10 % n1;
  for (long long i = n0 + n1; i >= n0 + n1 - cyc + 1; i--) s[i] = 0;
  s[n0 + n1 - cyc - n1 + rem] = 0;
  long long j = 0;
  for (long long i = n0 + n1; i >= 1; i--) {
    if (j == n1) s[i] = 0;
    if (j < n1 && s[i] == 1) j++;
  }
  for (long long i = 1; i <= n0 + n1; i++) cout << s[i];
  cout << endl;
  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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long fh = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= fh;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
long long s[1000005];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a00, a01, a10, a11;
  cin >> a00 >> a01 >> a10 >> a11;
  if (!a00 && !a01 && !a10 && !a11) {
    cout << 0 << endl;
    return 0;
  }
  if ((long long)(sqrt(1 + 8 * a00)) * (long long)(sqrt(1 + 8 * a00)) !=
      1 + 8 * a00) {
    cout << "Impossible" << endl;
    return 0;
  }
  if ((long long)(sqrt(1 + 8 * a11)) * (long long)(sqrt(1 + 8 * a11)) !=
      1 + 8 * a11) {
    cout << "Impossible" << endl;
    return 0;
  }
  long long n0 = (1 + sqrt(1 + 8 * a00)) / 2, n1 = (1 + sqrt(1 + 8 * a11)) / 2;
  if (a00 == 0) {
    if (!a01 && !a10) {
      for (long long i = 1; i <= n1; i++) cout << 1;
      cout << endl;
      return 0;
    }
  }
  if (a11 == 0) {
    if (!a01 && !a10) {
      for (long long i = 1; i <= n0; i++) cout << 0;
      cout << endl;
      return 0;
    }
  }
  if (n0 * n1 != a01 + a10) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (long long i = 1; i <= n0 + n1; i++) s[i] = 1;
  long long cyc = a10 / n1, rem = a10 % n1;
  for (long long i = n0 + n1; i >= n0 + n1 - cyc + 1; i--) s[i] = 0;
  s[n0 + n1 - cyc - n1 + rem] = 0;
  long long j = 0;
  for (long long i = n0 + n1; i >= 1; i--) {
    if (j == n1) s[i] = 0;
    if (j < n1 && s[i] == 1) j++;
  }
  for (long long i = 1; i <= n0 + n1; i++) cout << s[i];
  cout << endl;
  return 0;
}

```