Problem: 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:

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The prettiness of s equals 

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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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool isVowel(char ch) {
  return ch == 'A' or ch == 'E' or ch == 'I' or ch == 'O' or ch == 'U' or
         ch == 'Y';
}
int const MX = 5e5 + 100;
long long dp[MX];
long long cum[MX];
long long sub(int i, int j) { return cum[j] - cum[i - 1]; }
int main() {
  string s;
  cin >> s;
  int n = s.size();
  cum[1] = isVowel(s[0]);
  for (int i = 2; i <= s.size(); i++) {
    cum[i] = cum[i - 1] + isVowel(s[i - 1]);
  }
  dp[1] = sub(1, s.size());
  for (int i = 2; 2 * i <= s.size(); i++) {
    dp[i] = dp[i - 1] + sub(i, (int)s.size() - i + 1);
  }
  dp[n] = sub(1, s.size());
  for (int i = n - 1; i * 2 > s.size(); i--) {
    dp[i] = dp[i + 1] + sub(n - i + 1, i);
  }
  double ans = 0.0;
  for (int i = 1; i <= n; i++) {
    ans += 1.0 * dp[i] / (n - i + 1);
  }
  printf("%.15f\n", ans);
  return 0;
}