Good Bye 2017-A. New Year and Counting Cards

A. New Year and Counting Cards

time limit per test

memory limit per test

input

output

n

You know that each card has a lowercase English letter on one side and a digit on the other.

Currently, your friend has laid out the cards on a table so only one side of each card is visible.

a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.

a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.

To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.

Input

s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s

Output

Print a single integer, the minimum number of cards you must turn over to verify your claim.

Examples

input

ee

output

2

input

z

output

0

input

0ay1

output

2

Note

In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.

In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.

In the third sample, we need to flip the second and fourth cards.

熬夜打到2点钟，终于让rating升了一点……

这题很水，只要是奇数或元音就加一。

Code：

#include<bits/stdc++.h>
using namespace std;
int main()
{
  char s[55];int ans=0;
  scanf("%s",&s);
  for(int i=0;i<strlen(s);i++)
    if(s[i]=='1'||s[i]=='3'||s[i]=='5'||s[i]=='7'||s[i]=='9'||s[i]=='a'||s[i]=='o'||s[i]=='e'||s[i]=='i'||s[i]=='u')
      ans++;
  cout<<ans;
  return 0;
}