Your friend has *n* cards.

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.

You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.

For example, if a card has '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.
The first and only line of input will contain a string *s* (1 ≤ |*s*| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
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.
ee
2
z
0
0ay1
2
0abcdefghijklmnopqrstuvwxyz1234567896
10
0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b
18
01234567890123456789012345678901234567890123456789
25
qwertyuioplkjhgfdsazxcvbnmqwertyuioplkjhgfdsazxcvb
10
cjw2dwmr10pku4yxohe0wglktd
4
6z2tx805jie8cfybwtfqvmlveec3iak5z5u3lu62vbxyqht6
13
kaq7jyialrfp4ilkni90eq8v3amcbygon7py0hb8z26fbl8ss1
13
hpwn50zgbmct80k9rizjqg40nycgs0acwikjqt11nr6m61krfs
8
l3rw91a4m25l8iytxyeuixsegzcbm4h41ornf3pixkrmwznrzc
14
2222
0
13579
5
1
1
0
0
a
1
y
0
w
0
oo
2
oy
1
yo
1
yy
0
a0
1
a9
2
y0
0
y7
1
0a
1
3a
2
06
0
07
1
70
1
77
2
13570
4
0000000000011111
5
1357
4
uuuuuuuuuuuuuuuuuu
18
gabieurat
5
a=input()
vowel="aeiou"
num="02468"
count=0
for i in range(0,len(a)):
    if a[i] in vowel or a[i] in num:
        count+=1
print(count)