Divisibility by 25 CodeForces - 988E

You are given an integer nn from 11 to 10181018 without leading zeroes.

In one move you can swap any two adjacent digits in the given number in such a way that the resulting number will not contain leading zeroes. In other words, after each move the number you have cannot contain any leading zeroes.

What is the minimum number of moves you have to make to obtain a number that is divisible by 2525? Print -1 if it is impossible to obtain a number that is divisible by 2525.

Input

The first line contains an integer nn (1≤n≤10181≤n≤1018). It is guaranteed that the first (left) digit of the number nn is not a zero.

Output

If it is impossible to obtain a number that is divisible by 2525, print -1. Otherwise print the minimum number of moves required to obtain such number.

Note that you can swap only adjacent digits in the given number.

Examples

Input

5071

Output

4

Input

705

Output

1

Input

1241367

Output

-1

Note

In the first example one of the possible sequences of moves is 5071 →→ 5701 →→ 7501 →→ 7510 →→ 7150.

题意：一个大数，只能移动相邻的两位，问移动几次后可以除以25后没有余数

思路：一个模拟，末尾是00，25，50，75就可以了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define INF 0x3f3f3f3f
const long long int maxn=2e5+3;
string str;
int ans;
void solve(char a,char b)
{
    string s=str;
    int cnt1=s.rfind(a),cnt2=s.rfind(b);
    if(a=='0' && b=='0')
        cnt1=s.rfind(a,cnt1-1);
    if(cnt1==string::npos || cnt2==string::npos)
        return ;
    int cnt=0;
    if (cnt1 > cnt2)
    {
        cnt++;
        swap(cnt1, cnt2);
    }
    for (int i = cnt2; i+1 < s.size(); i++)
    {
        cnt++;
        swap(s[i], s[i+1]);
    }
    for (int i = cnt1; i+1 < s.size()-1; i++)
    {
        cnt++;
        swap(s[i], s[i+1]);
    }
    cnt += find_if(s.begin(), s.end(), [](char c){return c != '0';}) - s.begin();
    ans = min(ans, cnt);
}
    

int main()
{
    cin>>str;
    ans=INF;
    solve('0','0');
    solve('2','5');
    solve('5','0');
    solve('7','5');
    if(ans == INF)
        ans=-1;
    cout<<ans<<endl;
    return 0;
}