CF509E：Pretty Song（规律，前缀和）

E. Pretty Song

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

Output

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

Examples

input

IEAIAIO

output

28.0000000

input

BYOB

output

5.8333333

input

YISVOWEL

output

17.0500000

Note

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.

大意：给定一个字符串s，计算其中各子串元音字母所占比例，输出比例之和。

假定sum[i]表示1~i中元音字母的个数，len表示字符串总长度，字符串从下标1开始。

对于长度为1的子串，每个元音字母都要用1次，即(sum[len] -sum[0]) /  1，

对于长度为2的子串，s[2]到第s[len-1]的元音字母都要用2次，而第s[1]和s[len]个位置的元音字母只需用1次，即( (sum[len] -sum[0])  + (sum[len-1] - sum[1]) ) /  2,

对于长度为3的子串，s[3]到第s[len-2]的元音字母都要用3次，s[2]到第s[len-1]的元音字母都要用2次，而第s[1]和s[len]个位置的元音字母只需用1次，即( (sum[len] -sum[0])  + (sum[len-1] - sum[1])  + (sum[len-2] - sum[2]) )  /  3，

依此类推，各个长度的结果加起来就是答案。

# include <stdio.h># include <string.h># define MAXN 500000int sum[MAXN+3];int main(){    char c;    int len=0, i;    double result=0, ans=0;    memset(sum, 0, sizeof(sum));    while((c = getchar()) != '\n')    {        ++len;        if(c=='A'||c=='E'||c=='I'||c=='O'||c=='Y'||c=='U')            ++sum[len];        sum[len] += sum[len-1];    }    for(i=0; i<len; ++i)    {        ans += sum[len-i] - sum[i];        result += ans / (i+1);    }    printf("%.7lf\n",result);    return 0;}