#include <bits/stdc++.h>
using namespace std;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
char s[(1 << 18)];
long long Z, z, O, o, I, J, u, v;
bool sl() {
  while (I * (I - 1) / 2 < Z) ++I;
  while (J * (J - 1) / 2 < O) ++J;
  if (!I && (z || o)) ++I;
  if (!J && (o || z)) ++J;
  if (I * (I - 1) / 2 ^ Z || J * (J - 1) / 2 ^ O) return 0;
  for (int i(0); i < I; i++) s[i] = 48;
  for (int i(0); i < J; i++) s[i + I] = 49;
  u = 0, v = I, z -= I * J;
  if (z > 0 || z ^ -o) return 0;
  while (z < 0) {
    if (z < -I)
      swap(s[u++], s[v++]), z += I, o -= I;
    else
      while (z < 0) swap(s[v], s[v - 1]), --v, ++z, --o;
  }
  return 1;
}
int main(void) {
  scanf("%lld%lld%lld%lld", &Z, &z, &o, &O);
  if (!(z | Z | o | O)) return puts("0"), 0;
  if (!(Z | O) && o == 1 && !z) return puts("10"), 0;
  if (!(Z | O) && z == 1 && !o) return puts("01"), 0;
  puts(sl() ? s : "Impossible");
  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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
char s[(1 << 18)];
long long Z, z, O, o, I, J, u, v;
bool sl() {
  while (I * (I - 1) / 2 < Z) ++I;
  while (J * (J - 1) / 2 < O) ++J;
  if (!I && (z || o)) ++I;
  if (!J && (o || z)) ++J;
  if (I * (I - 1) / 2 ^ Z || J * (J - 1) / 2 ^ O) return 0;
  for (int i(0); i < I; i++) s[i] = 48;
  for (int i(0); i < J; i++) s[i + I] = 49;
  u = 0, v = I, z -= I * J;
  if (z > 0 || z ^ -o) return 0;
  while (z < 0) {
    if (z < -I)
      swap(s[u++], s[v++]), z += I, o -= I;
    else
      while (z < 0) swap(s[v], s[v - 1]), --v, ++z, --o;
  }
  return 1;
}
int main(void) {
  scanf("%lld%lld%lld%lld", &Z, &z, &o, &O);
  if (!(z | Z | o | O)) return puts("0"), 0;
  if (!(Z | O) && o == 1 && !z) return puts("10"), 0;
  if (!(Z | O) && z == 1 && !o) return puts("01"), 0;
  puts(sl() ? s : "Impossible");
  return 0;
}

```