#include <bits/stdc++.h>
using namespace std;
long long get(long long x) {
  for (long long i = 0; i * (i - 1) / 2 <= x; i++) {
    if (i * (i - 1) / 2 == x) return i;
  }
  cout << "Impossible" << endl;
  exit(0);
}
long long cnt[1000001];
int main() {
  long long a, b, c, d;
  cin >> a >> b >> c >> d;
  long long ax = a, bx = b, cx = c, dx = d;
  a = get(a);
  d = get(d);
  if (!a) a = (b > 0 || c > 0);
  if (!d) d = (b > 0 || c > 0);
  if (!a && !d) {
    if (!c && !b) {
      cout << 1 << endl;
      return 0;
    }
    cout << "Impossible" << endl;
    exit(0);
  }
  long long t = b;
  for (long long i = 1; i <= a; i++) {
    cnt[min(d, t)]++;
    t -= min(d, t);
  }
  vector<long long> ans;
  for (long long i = 0; i <= d; i++) {
    if (i) ans.push_back(1);
    while (cnt[i]--) ans.push_back(0);
  }
  int z = 0, o = 0;
  for (int x : ans) {
    if (x) {
      dx -= o;
      cx -= z;
      ++o;
    } else {
      ax -= z;
      bx -= o;
      ++z;
    }
  }
  if (ax || bx || cx || dx) {
    cout << "Impossible" << endl;
    exit(0);
  }
  reverse(ans.begin(), ans.end());
  for (int x : ans) cout << x;
}

### 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;
long long get(long long x) {
  for (long long i = 0; i * (i - 1) / 2 <= x; i++) {
    if (i * (i - 1) / 2 == x) return i;
  }
  cout << "Impossible" << endl;
  exit(0);
}
long long cnt[1000001];
int main() {
  long long a, b, c, d;
  cin >> a >> b >> c >> d;
  long long ax = a, bx = b, cx = c, dx = d;
  a = get(a);
  d = get(d);
  if (!a) a = (b > 0 || c > 0);
  if (!d) d = (b > 0 || c > 0);
  if (!a && !d) {
    if (!c && !b) {
      cout << 1 << endl;
      return 0;
    }
    cout << "Impossible" << endl;
    exit(0);
  }
  long long t = b;
  for (long long i = 1; i <= a; i++) {
    cnt[min(d, t)]++;
    t -= min(d, t);
  }
  vector<long long> ans;
  for (long long i = 0; i <= d; i++) {
    if (i) ans.push_back(1);
    while (cnt[i]--) ans.push_back(0);
  }
  int z = 0, o = 0;
  for (int x : ans) {
    if (x) {
      dx -= o;
      cx -= z;
      ++o;
    } else {
      ax -= z;
      bx -= o;
      ++z;
    }
  }
  if (ax || bx || cx || dx) {
    cout << "Impossible" << endl;
    exit(0);
  }
  reverse(ans.begin(), ans.end());
  for (int x : ans) cout << x;
}

```