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
#include <bits/stdc++.h>
using namespace std;
int calc(int x) {
  if (x == 0) return 0;
  for (int i = 1; i <= 1000000; i++) {
    if (i * (i - 1) == x * 2) return i;
  }
  return -1;
}
int main() {
  int a00, a01, a10, a11;
  scanf("%d%d%d%d", &a00, &a01, &a10, &a11);
  int x = calc(a00), y = calc(a11);
  if (!x && (a01 || a10)) x = 1;
  if (!y && (a01 || a10)) y = 1;
  if (x == -1 || y == -1 || x * y != a01 + a10) {
    puts("Impossible");
    return 0;
  }
  if (x == 0 && y == 0) {
    puts("0");
    return 0;
  }
  if (x == 0) {
    for (int i = 1; i <= y; i++) putchar('1');
    printf("\n");
    return 0;
  }
  if (y == 0) {
    for (int i = 1; i <= x; i++) putchar('0');
    printf("\n");
    return 0;
  }
  int u = a01 / y, v = y - a01 % y;
  for (int i = 1; i <= u; i++) putchar('0'), x--;
  for (int i = 1; i <= v; i++) putchar('1'), y--;
  if (x) putchar('0'), x--;
  for (int i = 1; i <= y; i++) putchar('1');
  for (int i = 1; i <= x; i++) putchar('0');
  printf("\n");
  return 0;
}
