Codeforces-962C-Make a Square（暴力二进制枚举）

C. Make a Square

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

You are given a positive integer $n$, written without leading zeroes (for example, the number 04 is incorrect).

In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.

Determine the minimum number of operations that you need to consistently apply to the given integer $n$ to make from it the square of some positive integer or report that it is impossible.

An integer $x$ is the square of some positive integer if and only if $x=y^2$ for some positive integer $y$.

Input

The first line contains a single integer $n$($1\le n\le 2\cdot {10}^9$). The number is given without leading zeroes.

Output

If it is impossible to make the square of some positive integer from $n$, print -1. In the other case, print the minimal number of operations required to do it.

Examples

Input

8314

Output

2

Input

625

Output

0

Input

333

Output

-1

Note

In the first example we should delete from $8314$the digits $3$ and$4$. After that$8314$ become equals to $81$, which is the square of the integer $9$.

In the second example the given $625$is the square of the integer $25$, so you should not delete anything.

In the third example it is impossible to make the square from $333$, so the answer is -1.

题意：稳最少去掉几个数字使剩下的数字是个完全平方数；

思路：最多10位，用二进制枚举，从1<<(len)-1开始，len是整数的位数，一直到0就可以遍历出每一种情况，二进制的1表示改为保留，0表示去掉，然后每个判断一下就行；

AC代码：

#include<bits/stdc++.h>
using namespace std;
int n,a[15];
int init(int x)
{
    int cnt=0;
    while(x){
        a[++cnt]=x%10;
        x/=10;
    }
    return cnt;///位数
}
int main()
{
    cin>>n;
    int len=init(n);
    int ans=100;
    for(int i=(1<<len)-1;i>=0;i--){
        int tmp=0,cnt=0,flag=1;
        for(int j=len-1;j>=0;j--){
            if((i>>j)&1){
                cnt++;///记录有多少个1
                tmp=tmp*10+a[j+1];
                if(tmp==0){
                    flag=0;
                    break;
                }
            }
        }
        if(tmp&&flag&&(int(sqrt(tmp))*int(sqrt(tmp))==tmp)){
            ans=min(ans,len-cnt);
        }
    }
    if(ans!=100) printf("%d\n",ans);
    else  printf("-1\n");
    return 0;
}