#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int Left[N];
int ans[N];
int a, b, c, d;
int getval(int x) {
  int l = 1, r = 45000;
  while (l <= r) {
    int mid = (l + r) >> 1;
    int t = mid * (mid - 1) / 2;
    if (t == x) return mid;
    if (t < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return -1;
}
int main() {
  int n;
  while (cin >> a >> b >> c >> d) {
    int x, y;
    if (a == 0)
      x = 1;
    else
      x = getval(a);
    if (d == 0)
      y = 1;
    else
      y = getval(d);
    if (x == 1) {
      if (b == 0 && c == 0 && y >= 0) {
        while (y--) printf("1");
        printf("\n");
        continue;
      }
    }
    if (y == 1) {
      if (b == 0 && c == 0 && x >= 0) {
        while (x--) printf("0");
        printf("\n");
        continue;
      }
    }
    if (x == -1 || y == -1 || (x * y != b + c)) {
      puts("Impossible");
      continue;
    }
    int flag = 1;
    memset(Left, 0, sizeof(Left));
    for (int i = x; i >= 1; i--) {
      if (c > y && i == 1) {
        flag = 0;
        break;
      }
      if (c >= y) {
        Left[i] = y;
        c -= y;
      } else {
        Left[i] = c;
        break;
      }
    }
    if (flag == 0) {
      puts("Impossible");
      continue;
    }
    Left[x + 1] = y;
    int t = x + y;
    for (int i = x; i >= 1; i--) {
      for (int j = 1; j <= Left[i + 1] - Left[i]; j++) ans[t--] = 1;
      ans[t--] = 0;
    }
    for (int j = 1; j <= Left[1]; j++) ans[t--] = 1;
    for (int i = 1; i <= x + y; i++) printf("%d", ans[i]);
    printf("\n");
  }
  return 0;
}

### 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;
const int N = 1e5 + 5;
int Left[N];
int ans[N];
int a, b, c, d;
int getval(int x) {
  int l = 1, r = 45000;
  while (l <= r) {
    int mid = (l + r) >> 1;
    int t = mid * (mid - 1) / 2;
    if (t == x) return mid;
    if (t < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return -1;
}
int main() {
  int n;
  while (cin >> a >> b >> c >> d) {
    int x, y;
    if (a == 0)
      x = 1;
    else
      x = getval(a);
    if (d == 0)
      y = 1;
    else
      y = getval(d);
    if (x == 1) {
      if (b == 0 && c == 0 && y >= 0) {
        while (y--) printf("1");
        printf("\n");
        continue;
      }
    }
    if (y == 1) {
      if (b == 0 && c == 0 && x >= 0) {
        while (x--) printf("0");
        printf("\n");
        continue;
      }
    }
    if (x == -1 || y == -1 || (x * y != b + c)) {
      puts("Impossible");
      continue;
    }
    int flag = 1;
    memset(Left, 0, sizeof(Left));
    for (int i = x; i >= 1; i--) {
      if (c > y && i == 1) {
        flag = 0;
        break;
      }
      if (c >= y) {
        Left[i] = y;
        c -= y;
      } else {
        Left[i] = c;
        break;
      }
    }
    if (flag == 0) {
      puts("Impossible");
      continue;
    }
    Left[x + 1] = y;
    int t = x + y;
    for (int i = x; i >= 1; i--) {
      for (int j = 1; j <= Left[i + 1] - Left[i]; j++) ans[t--] = 1;
      ans[t--] = 0;
    }
    for (int j = 1; j <= Left[1]; j++) ans[t--] = 1;
    for (int i = 1; i <= x + y; i++) printf("%d", ans[i]);
    printf("\n");
  }
  return 0;
}

```