AIM Tech Round 3 (Div. 2)D. Recover the String（贪心+字符串）

D. 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 exceed109.

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

 比赛时，一直纠结怎么判断Impossible和构造字符串，胡乱交了一发，wa了，好SB。

当然先是利用a00和a11求1和0的字符串个数g1和g0，如果a00或a11等于1，这时候还要根据a10和a01判断a00或者a11是等于1还是0.

在求完a00和a11的个数后，判断合不合理，C(g0,2)=a00, C(g1,2)=a11, C(g1+g0,2) = a00+a01+a10+a11即成立。

然后就是构造字符串了。

设初始的字符串个数为0000001111111......

然后就是贪心的构造字符串了。考虑该输出某一位时，如果在这一位时，a10>=g0,那么我可以把1挪到最前面，输出1，这时候后面剩下的字符串需要构造的a10-=g0,剩余的1的个数g1--。否则直接输出0，后面剩余字符串需要构造a01-=g1,剩余的0的个数g0--.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
int main()
{
    ll a00,a01,a10,a11;
    scanf("%I64d%I64d%I64d%I64d",&a00,&a01,&a10,&a11);
    ll g0,g1;
    g0 = (1+sqrt(1+8*a00))/2;
    g1 = (1+sqrt(1+8*a11))/2;
    if(a00+a01+a10+a11==0)
    {
         printf("0
");
         return 0;
    }
    if(g0==1||g1==1)
    {
        if(g0==1)
        {
            if(a01||a10) g0 = 1;
            else g0 = 0;
        }
        if(g1==1)
        {
            if(a01||a10) g1 = 1;
            else g1 = 0;
        }
    }
    if(g0*(g0-1)!=2*a00||g1*(g1-1)!=2*a11)
    {
        printf("Impossible
");
        return 0;
    }
    ll gz = g0+g1;
    if(gz*(gz-1)/2!=(a00+a01+a10+a11))
    {
        printf("Impossible
");
        return 0;
    }
    while(g0+g1)
    {
        if(a10>=g0)
        {
            printf("1");
            a10 -= g0;
            g1--;
        }
        else
        {
            printf("0");
            a01 -= g1;
            g0--;
        }
    }
    printf("
");
    return 0;
}