#include <bits/stdc++.h>
using namespace std;
long long n, m, u, v, q, r, ql, qr, k, l, s, w, z, x, y, d, p, h, b, c;
const int N = 2e5 + 500;
const int NN = N + N + N;
const int LN = 21;
const long long mod = 1e9 + 7;
const long long INF = 1LL << 52LL;
long long zz, zo, oz, oo;
string sol, str;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> zz >> zo >> oz >> oo;
  long long x = 0, y = 0;
  while (x * (x - 1) / 2 < zz) x++;
  while (y * (y - 1) / 2 < oo) y++;
  if (zz == 0 && (zo > 0 || oz > 0)) x = 1;
  if (oo == 0 && (oz > 0 || zo > 0)) y = 1;
  if (y * (y - 1) / 2 != oo) {
    cout << "Impossible\n";
    return 0;
  }
  if (x * (x - 1) / 2 != zz) {
    cout << "Impossible\n";
    return 0;
  }
  long long z = x + y;
  if (z * (z - 1) / 2 != zz + zo + oz + oo) {
    cout << "Impossible\n";
    return 0;
  }
  if (zo + oz != x * y) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i = 0; i < x; ++i) sol.push_back('0');
  while (zo >= x && x != 0) {
    sol.push_back('1');
    zo -= x;
    y--;
  }
  if (zo > 0) {
    sol.insert(zo, "1");
    y--;
  }
  for (int i = 0; i < y; ++i) str.push_back('1');
  str = str + sol;
  if (str == "") str = "1";
  cout << str << '\n';
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 n, m, u, v, q, r, ql, qr, k, l, s, w, z, x, y, d, p, h, b, c;
const int N = 2e5 + 500;
const int NN = N + N + N;
const int LN = 21;
const long long mod = 1e9 + 7;
const long long INF = 1LL << 52LL;
long long zz, zo, oz, oo;
string sol, str;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> zz >> zo >> oz >> oo;
  long long x = 0, y = 0;
  while (x * (x - 1) / 2 < zz) x++;
  while (y * (y - 1) / 2 < oo) y++;
  if (zz == 0 && (zo > 0 || oz > 0)) x = 1;
  if (oo == 0 && (oz > 0 || zo > 0)) y = 1;
  if (y * (y - 1) / 2 != oo) {
    cout << "Impossible\n";
    return 0;
  }
  if (x * (x - 1) / 2 != zz) {
    cout << "Impossible\n";
    return 0;
  }
  long long z = x + y;
  if (z * (z - 1) / 2 != zz + zo + oz + oo) {
    cout << "Impossible\n";
    return 0;
  }
  if (zo + oz != x * y) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i = 0; i < x; ++i) sol.push_back('0');
  while (zo >= x && x != 0) {
    sol.push_back('1');
    zo -= x;
    y--;
  }
  if (zo > 0) {
    sol.insert(zo, "1");
    y--;
  }
  for (int i = 0; i < y; ++i) str.push_back('1');
  str = str + sol;
  if (str == "") str = "1";
  cout << str << '\n';
  return 0;
}

```