When Sasha was studying in the seventh grade, he started listening to music a lot. In order to evaluate which songs he likes more, he introduced the notion of the song's prettiness. The title of the song is a word consisting of uppercase Latin letters. The prettiness of the song is the prettiness of its title.

Let's define the simple prettiness of a word as the ratio of the number of vowels in the word to the number of all letters in the word.

Let's define the prettiness of a word as the sum of simple prettiness of all the substrings of the word.

More formally, let's define the function vowel(c) which is equal to 1, if c is a vowel, and to 0 otherwise. Let si be the i-th character of string s, and si..j be the substring of word s, staring at the i-th character and ending at the j-th character (sisi + 1... sj, i ≤ j).

Then the simple prettiness of s is defined by the formula:

The prettiness of s equals

Find the prettiness of the given song title.

We assume that the vowels are I, E, A, O, U, Y.

The input contains a single string s (1 ≤ |s| ≤ 5·105) — the title of the song.

Print the prettiness of the song with the absolute or relative error of at most 10 - 6.

In the first sample all letters are vowels. The simple prettiness of each substring is 1. The word of length 7 has 28 substrings. So, the prettiness of the song equals to 28.

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<iostream>

#include<cstdio>

#include<cmath>

#include<string>

#include<queue>

#include<algorithm>

#include<stack>

#include<cstring>

#include<vector>

#include<list>

#include<set>

#include<map>

using namespace std;

#define ll long long

#define pi (4*atan(1.0))

#define mk make_pair

#define eps 1e-7

#define bug(x)  cout<<"bug"<<x<<endl;

const int N=5e5+,M=1e6+,inf=;

const ll INF=1e18+,mod=;

///   数组大小

char ch[]={'I','E','A','O','U','Y'},s[N];

double sum1[N],sum2[N],sum3[N];

int check(char a)

{

    for(int i=;i<;i++)

        if(a==ch[i])return ;

    return ;

}

int main()

{

    scanf("%s",s+);

    int n=strlen(s+);

    for(int i=;i<=n;i++)

        sum1[i]=sum1[i-]+(1.0*/i);

    for(int i=;i<=n;i++)

        sum2[i]=sum2[i-]+sum1[i];

    for(int i=n,j=;i>=;i--,j++)

        sum3[j]=sum3[j-]+sum1[n]-sum1[i-];

    double ans=0.0;

    for(int i=;i<=n;i++)

    {

        if(check(s[i]))

        {

            int l=i;

            int r=n-i+;

            ans+=1.0*sum1[n]*l;

            ans-=sum2[l-];

            ans-=sum3[l-];

        }

        //cout<<ans<<endl;

    }

    printf("%f\n",ans);

    return ;

}

When Sasha was studying in the seventh grade, he started listening to music a lot. In order to evaluate which songs he likes more, he introduced the notion of the song's prettiness. The title of the song is a word consisting of uppercase Latin letters. The prettiness of the song is the prettiness of its title.

Let's define the simple prettiness of a word as the ratio of the number of vowels in the word to the number of all letters in the word.

Let's define the prettiness of a word as the sum of simple prettiness of all the substrings of the word.

More formally, let's define the function vowel(c) which is equal to 1, if c is a vowel, and to 0 otherwise. Let si be the i-th character of string s, and si..j be the substring of word s, staring at the i-th character and ending at the j-th character (sisi + 1... sj, i ≤ j).

Then the simple prettiness of s is defined by the formula:

The prettiness of s equals

Find the prettiness of the given song title.

We assume that the vowels are I, E, A, O, U, Y.

The input contains a single string s (1 ≤ |s| ≤ 5·105) — the title of the song.

Print the prettiness of the song with the absolute or relative error of at most 10 - 6.

In the first sample all letters are vowels. The simple prettiness of each substring is 1. The word of length 7 has 28 substrings. So, the prettiness of the song equals to 28.