#include <bits/stdc++.h>
using namespace std;
long long sq[100001];
int main() {
  sq[0] = 1;
  for (long long i = 1; i < 100001; i++) {
    sq[i] = i * (i - 1) / 2;
  }
  string res = "";
  int a, b, c, d;
  cin >> a >> b >> c >> d;
  if (a == 0 and b == 0 and c == 0 and d == 0) {
    cout << 0 << endl;
    return 0;
  }
  int z = lower_bound(sq, sq + 100001, a) - sq;
  int o = lower_bound(sq, sq + 100001, d) - sq;
  if (z != 0 and sq[z] != a or o != 0 and sq[o] != d) {
    cout << "Impossible" << endl;
    return 0;
  }
  if (a == 0 and b == 0 and c == 0) {
    string res = string(o, '1');
    cout << res << endl;
    return 0;
  }
  if (b == 0 and c == 0 and d == 0) {
    string res = string(z, '0');
    cout << res << endl;
    return 0;
  }
  if (o == 0) o = 1;
  if (z == 0) z = 1;
  if (z * o != b + c) {
    cout << "Impossible" << endl;
    return 0;
  }
  if (b >= c) {
    int need = b;
    int opt = need / o;
    string q = string(opt, '0');
    res += q;
    z -= opt;
    need -= opt * o;
    q = string(o - need, '1');
    res += q;
    if (z > 0) res += '0';
    z--;
    q = string(need, '1');
    res += q;
    q = string(max(0, z), '0');
    res += q;
    cout << res << endl;
    return 0;
  } else {
    int need = c;
    int opt = need / z;
    string q = string(opt, '1');
    res += q;
    o -= opt;
    need -= opt * z;
    q = string(z - need, '0');
    res += q;
    if (o > 0) res += '1';
    o--;
    q = string(need, '0');
    res += q;
    q = string(max(0, o), '1');
    res += q;
    cout << res << endl;
    return 0;
  }
  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;
long long sq[100001];
int main() {
  sq[0] = 1;
  for (long long i = 1; i < 100001; i++) {
    sq[i] = i * (i - 1) / 2;
  }
  string res = "";
  int a, b, c, d;
  cin >> a >> b >> c >> d;
  if (a == 0 and b == 0 and c == 0 and d == 0) {
    cout << 0 << endl;
    return 0;
  }
  int z = lower_bound(sq, sq + 100001, a) - sq;
  int o = lower_bound(sq, sq + 100001, d) - sq;
  if (z != 0 and sq[z] != a or o != 0 and sq[o] != d) {
    cout << "Impossible" << endl;
    return 0;
  }
  if (a == 0 and b == 0 and c == 0) {
    string res = string(o, '1');
    cout << res << endl;
    return 0;
  }
  if (b == 0 and c == 0 and d == 0) {
    string res = string(z, '0');
    cout << res << endl;
    return 0;
  }
  if (o == 0) o = 1;
  if (z == 0) z = 1;
  if (z * o != b + c) {
    cout << "Impossible" << endl;
    return 0;
  }
  if (b >= c) {
    int need = b;
    int opt = need / o;
    string q = string(opt, '0');
    res += q;
    z -= opt;
    need -= opt * o;
    q = string(o - need, '1');
    res += q;
    if (z > 0) res += '0';
    z--;
    q = string(need, '1');
    res += q;
    q = string(max(0, z), '0');
    res += q;
    cout << res << endl;
    return 0;
  } else {
    int need = c;
    int opt = need / z;
    string q = string(opt, '1');
    res += q;
    o -= opt;
    need -= opt * z;
    q = string(z - need, '0');
    res += q;
    if (o > 0) res += '1';
    o--;
    q = string(need, '0');
    res += q;
    q = string(max(0, o), '1');
    res += q;
    cout << res << endl;
    return 0;
  }
  return 0;
}

```