#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
string toBin(long long a) {
  string res = "";
  while (a) {
    res += char((a & 1) + '0');
    a >>= 1;
  }
  reverse(res.begin(), res.end());
  return res;
}
const int mod = 1e9 + 7;
long long expo(long long base, long long pow) {
  long long ans = 1;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans;
}
long long inv(long long x) { return expo(x, mod - 2); }
bool isPal(string ss) {
  int len = ss.length();
  for (int i = 0; i < len / 2; i++) {
    int comp = len - i - 1;
    if (ss[i] != ss[comp]) return false;
  }
  return true;
}
double pi = 3.141592653589793238462643;
double error = 0.0000001;
const int M = 100001;
vector<long long> get2(long long ex) {
  if (ex == 0) return {0, 1};
  long long x = 1;
  vector<long long> vv;
  while ((x * (x - 1)) / 2 < ex) x++;
  if ((x * (x - 1)) / 2 == ex) vv.push_back(x);
  return vv;
}
int main() {
  long long a, b, c, d;
  scanf("%lld", &(a));
  scanf("%lld", &(b));
  scanf("%lld", &(c));
  scanf("%lld", &(d));
  vector<long long> on = get2(a);
  vector<long long> tw = get2(d);
  for (int i = 0; i < on.size(); i++) {
    for (int j = 0; j < tw.size(); j++) {
      long long zez = on[i];
      long long onz = tw[j];
      if (zez + onz == 0) continue;
      if (zez * onz != b + c) continue;
      vector<int> ans;
      long long tot = zez + onz;
      for (int i = 0; i < tot; i++) {
        if (b >= onz) {
          b -= onz;
          zez--;
          ans.push_back(0);
        } else {
          c -= zez;
          onz--;
          ans.push_back(1);
        }
      }
      for (long long i = 0; i < tot; i++) {
        printf("%d", ans[i]);
      }
      return 0;
    }
  }
  cout << "Impossible";
}

### Prompt

Please create a solution in CPP 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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
string toBin(long long a) {
  string res = "";
  while (a) {
    res += char((a & 1) + '0');
    a >>= 1;
  }
  reverse(res.begin(), res.end());
  return res;
}
const int mod = 1e9 + 7;
long long expo(long long base, long long pow) {
  long long ans = 1;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans;
}
long long inv(long long x) { return expo(x, mod - 2); }
bool isPal(string ss) {
  int len = ss.length();
  for (int i = 0; i < len / 2; i++) {
    int comp = len - i - 1;
    if (ss[i] != ss[comp]) return false;
  }
  return true;
}
double pi = 3.141592653589793238462643;
double error = 0.0000001;
const int M = 100001;
vector<long long> get2(long long ex) {
  if (ex == 0) return {0, 1};
  long long x = 1;
  vector<long long> vv;
  while ((x * (x - 1)) / 2 < ex) x++;
  if ((x * (x - 1)) / 2 == ex) vv.push_back(x);
  return vv;
}
int main() {
  long long a, b, c, d;
  scanf("%lld", &(a));
  scanf("%lld", &(b));
  scanf("%lld", &(c));
  scanf("%lld", &(d));
  vector<long long> on = get2(a);
  vector<long long> tw = get2(d);
  for (int i = 0; i < on.size(); i++) {
    for (int j = 0; j < tw.size(); j++) {
      long long zez = on[i];
      long long onz = tw[j];
      if (zez + onz == 0) continue;
      if (zez * onz != b + c) continue;
      vector<int> ans;
      long long tot = zez + onz;
      for (int i = 0; i < tot; i++) {
        if (b >= onz) {
          b -= onz;
          zez--;
          ans.push_back(0);
        } else {
          c -= zez;
          onz--;
          ans.push_back(1);
        }
      }
      for (long long i = 0; i < tot; i++) {
        printf("%d", ans[i]);
      }
      return 0;
    }
  }
  cout << "Impossible";
}

```