#include <bits/stdc++.h>
using namespace std;
long long poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const long long MAXN = 1e6 + 10;
const long long INF = 8e18;
const long long MOD = 1e9 + 7;
long long ftn(long long x) {
  long long i = 0;
  while (i * (i - 1) / 2 < x) i++;
  if (i * (i - 1) / 2 > x) return -1;
  return i;
}
long long a00, a01, a10, a11;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  cin >> a00 >> a01 >> a10 >> a11;
  long long o = ftn(a11), z = ftn(a00);
  if (o == 0 && (a01 || a10)) o++;
  if (z == 0 && (a01 || a10)) z++;
  long long n = o * z;
  if (a01 == 0 && a10 == 0 && a00 == 0 && a11 == 0) return cout << 1 << '\n', 0;
  if (o == -1 || z == -1 || a01 + a10 != n)
    return cout << "Impossible" << '\n', 0;
  long long x = 0, y = 0;
  if (z > 0) x = a10 / z, y = a10 % z;
  if (x + min(y, 1ll) > o) return cout << "Impossible" << '\n', 0;
  for (int i = 0; i < x; i++) cout << '1', o--;
  for (int i = 0; i < z - y; i++) cout << '0';
  if (y) cout << '1', o--;
  for (int i = 0; i < y; i++) cout << '0';
  while (o > 0) cout << '1', o--;
  cout << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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 poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const long long MAXN = 1e6 + 10;
const long long INF = 8e18;
const long long MOD = 1e9 + 7;
long long ftn(long long x) {
  long long i = 0;
  while (i * (i - 1) / 2 < x) i++;
  if (i * (i - 1) / 2 > x) return -1;
  return i;
}
long long a00, a01, a10, a11;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  cin >> a00 >> a01 >> a10 >> a11;
  long long o = ftn(a11), z = ftn(a00);
  if (o == 0 && (a01 || a10)) o++;
  if (z == 0 && (a01 || a10)) z++;
  long long n = o * z;
  if (a01 == 0 && a10 == 0 && a00 == 0 && a11 == 0) return cout << 1 << '\n', 0;
  if (o == -1 || z == -1 || a01 + a10 != n)
    return cout << "Impossible" << '\n', 0;
  long long x = 0, y = 0;
  if (z > 0) x = a10 / z, y = a10 % z;
  if (x + min(y, 1ll) > o) return cout << "Impossible" << '\n', 0;
  for (int i = 0; i < x; i++) cout << '1', o--;
  for (int i = 0; i < z - y; i++) cout << '0';
  if (y) cout << '1', o--;
  for (int i = 0; i < y; i++) cout << '0';
  while (o > 0) cout << '1', o--;
  cout << '\n';
  return 0;
}

```