AIM Tech Round 3 (Div. 1)-B. Recover the String

原题链接

B. Recover the String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <map>#include <vector>#include <queue>#include <cmath>#define maxn 1000005#define MOD 1000000007#define INF 1e9using namespace std;typedef long long ll;char str[maxn];int main(){//freopen("in.txt", "r", stdin);ll a, b, c, d;scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d);a *= 2;d *= 2;ll k1 = sqrt(a);ll k2 = sqrt(d);if(k1 * (k1 + 1) != a || k2 * (k2 + 1) != d){puts("Impossible");return 0;}if(a == 0 && b == 0 && c == 0 && d == 0){puts("0");return 0;}if(a == 0 && b == 0 && c == 0 && d){for(int i = 1; i <= k2 + 1; i++) str[i] = '1';puts(str+1);return 0;}if(a && b == 0 && c == 0 && d == 0){for(int i  = 1; i <= k1 + 1; i++) str[i] = '0';puts(str+1);return 0;}if((k1+1) * (k2+1) != b + c){puts("Impossible");return 0;}k1++, k2++;ll cc = b / k2, e = b % k2;for(int i = 1; i <= k1 + k2; i++) str[i] = '0';for(int i = cc + 1; i <= cc + k2; i++) str[i] = '1';if(e){ str[cc+k2+1] = '1'; str[cc+k2+1-e] = '0';    }    puts(str+1);    return 0;}