codeforces 708B (数学构造)

For each string s consisting of characters '0' and '1' one can define four integersa₀₀, a₀₁, a₁₀ and a₁₁, where a_xy 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.

Example

Input

1 2 3 4

Output

Impossible

Input

1 2 2 1

Output

0110

题意：已知 a,b,c,d. 构造不大于1e6的字符串，使得其中子序列 00 ,01,10,11 的数量 = a,b,c,d。

已知，知道 00 11，的数目 可以知道 0,1的数量 x*(x-1) =a 快速求得正整数解得方法是 开方+1，然后相乘验证一下是否有解。

关于 01和10的数量，假设把0全放左边，1全放右边，得出01的数目就是 cnt1*cnt2 每次把1左移一位，10数量增加，01数量减少

#include <bits/stdc++.h>using namespace std;int num[1000010];int main(){int a,b,c,d;cin>>a>>b>>c>>d;int cnt1=sqrt(2*a)+1;int cnt2=sqrt(2*d)+1;if(cnt1*(cnt1-1)!=2*a||cnt2*(cnt2-1)!=2*d){puts("Impossible");return 0;}memset(num,0,sizeof(num));if(a+b+c+d==0){puts("0");return 0;}if(a+b+c==0){for(int i=1;i<=cnt2;i++)printf("1");return 0;}if(a&&b+c+d==0){for(int i=1;i<=cnt1;i++)printf("0");return 0;}if(cnt1*cnt2!=b+c){puts("Impossible");return 0;}int h=0;while(c){if(c>=cnt1){c-=cnt1;num[++h]=1;}else{++h;num[h+cnt1-c]=1;c=0;break;}}for(int i=1;i<=cnt2-h;i++)num[cnt1+cnt2+1-i]=1;for(int i=1;i<=cnt1+cnt2;i++)printf("%d",num[i] );puts("");}