Recover the String

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 ofsubsequences 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
分析：可以先求出0和1的个数，注意边界情况；
　　　注意到10和01的总个数就是0的个数*1的个数，所以只需构造出10即可；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t;
ll a,b,c,d,num1,num2;
string ans;
int main()
{
    int i,j;
    scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
    if(a+b+c+d==0)puts("0");                    //特判四个0的情况；
    else
    {
        int c1=1,c2=1;                          //求出1和0的个数；
        while(num1<a)num1+=c1,c1++;
        while(num2<d)num2+=c2,c2++;
        if(!a&&!b&&!c)c1=0;                     //特判0的个数为0时；
        if(!b&&!c&&!d)c2=0;                     //特判1的个数为0时；
        if(num1!=a||num2!=d||(ll)c1*c2!=(ll)b+c)puts("Impossible");         //判断是否可行；
        else
        {
            int j=0,now=0;
            while(j+c1<c)j+=c1,now++,c2--;                                  //在0里面不断插入1；
            rep(i,0,now-1)ans+='1';
            rep(i,now,now+c1-1)ans+='0';
            if(c>j)ans.insert(now+c1-c+j,"1"),c2--;                         //补10；
            rep(i,1,c2)ans+='1';                                            //末尾补1；
            cout<<ans<<endl;
        }
    }
    //system("Pause");
    return 0;
}