#include <bits/stdc++.h>
using namespace std;
const int N = 1000000;
int a[N + 20000];
int n, s[2][2];
void fail() {
  puts("Impossible");
  exit(0);
}
int cnt[2];
int main() {
  int i, j;
  long long sum = 0;
  for (i = 0; i < 2; i++)
    for (j = 0; j < 2; j++) {
      scanf("%d", &s[i][j]);
      sum += s[i][j];
    }
  if (!sum) {
    puts("1");
    return 0;
  }
  n = sqrt(sum * 2) + 1;
  if (1ll * n * (n - 1) / 2 != sum) {
    fail();
  }
  for (i = 0; i < 2; i++) {
    if (s[i][i] || s[!i][i] || s[i][!i])
      cnt[i] = sqrt(s[i][i] * 2) + 1;
    else
      cnt[i] = 0;
    if (1ll * cnt[i] * (cnt[i] - 1) / 2 != s[i][i]) {
      fail();
    }
  }
  if (cnt[0] + cnt[1] != n) {
    fail();
  }
  if (1ll * cnt[0] * cnt[1] != s[1][0] + s[0][1]) {
    fail();
  }
  if (s[0][1] > 1ll * cnt[0] * cnt[1] || s[1][0] > 1ll * cnt[0] * cnt[1]) {
    fail();
  }
  vector<int> lst;
  int target = s[0][1];
  for (i = 1; i <= cnt[1]; i++) {
    int tmp = min(cnt[0], target);
    lst.push_back(tmp);
    target -= tmp;
  }
  sort(lst.begin(), lst.end());
  int put0 = 0, top = 0;
  for (int i = 0; i < lst.size(); i++) {
    while (put0 < lst[i]) {
      put0++;
      a[++top] = 0;
    }
    a[++top] = 1;
  }
  while (top < n) a[++top] = 0;
  for (i = 1; i <= top; i++) printf("%d", a[i]);
  puts("");
}

### Prompt

Create a solution in Cpp for 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 = 1000000;
int a[N + 20000];
int n, s[2][2];
void fail() {
  puts("Impossible");
  exit(0);
}
int cnt[2];
int main() {
  int i, j;
  long long sum = 0;
  for (i = 0; i < 2; i++)
    for (j = 0; j < 2; j++) {
      scanf("%d", &s[i][j]);
      sum += s[i][j];
    }
  if (!sum) {
    puts("1");
    return 0;
  }
  n = sqrt(sum * 2) + 1;
  if (1ll * n * (n - 1) / 2 != sum) {
    fail();
  }
  for (i = 0; i < 2; i++) {
    if (s[i][i] || s[!i][i] || s[i][!i])
      cnt[i] = sqrt(s[i][i] * 2) + 1;
    else
      cnt[i] = 0;
    if (1ll * cnt[i] * (cnt[i] - 1) / 2 != s[i][i]) {
      fail();
    }
  }
  if (cnt[0] + cnt[1] != n) {
    fail();
  }
  if (1ll * cnt[0] * cnt[1] != s[1][0] + s[0][1]) {
    fail();
  }
  if (s[0][1] > 1ll * cnt[0] * cnt[1] || s[1][0] > 1ll * cnt[0] * cnt[1]) {
    fail();
  }
  vector<int> lst;
  int target = s[0][1];
  for (i = 1; i <= cnt[1]; i++) {
    int tmp = min(cnt[0], target);
    lst.push_back(tmp);
    target -= tmp;
  }
  sort(lst.begin(), lst.end());
  int put0 = 0, top = 0;
  for (int i = 0; i < lst.size(); i++) {
    while (put0 < lst[i]) {
      put0++;
      a[++top] = 0;
    }
    a[++top] = 1;
  }
  while (top < n) a[++top] = 0;
  for (i = 1; i <= top; i++) printf("%d", a[i]);
  puts("");
}

```