Problem:
Counting Characters

Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.

Input

A sentence in English is given in several lines.

Output

Prints the number of alphabetical letters in the following format:

a : The number of 'a'
b : The number of 'b'
c : The number of 'c'
.
.
z : The number of 'z'

Constraints

The number of characters in the sentence < 1200

Sample Input

This is a pen.

Sample Output

a : 1
b : 0
c : 0
d : 0
e : 1
f : 0
g : 0
h : 1
i : 2
j : 0
k : 0
l : 0
m : 0
n : 1
o : 0
p : 1
q : 0
r : 0
s : 2
t : 1
u : 0
v : 0
w : 0
x : 0
y : 0
z : 0

Solution in Ruby (judge verdict: Wrong Answer)

sent = gets.chomp
sent_len = sent.length

arr = sent.split("")

alpha = Array.new(26, 0)	# アルファベットの回数を記録する配列

for k in 0..sent_len - 1
	case arr[k]
		when "a"
			alpha[0] += 1
		when "b"
			alpha[1] += 1
		when "c"
			alpha[2] += 1
		when "d"
			alpha[3] += 1
		when "e"
			alpha[4] += 1
		when "f"
			alpha[5] += 1
		when "g"
			alpha[6] += 1
		when "h"
			alpha[7] += 1
		when "i"
			alpha[8] += 1
		when "j"
			alpha[9] += 1
		when "k"
			alpha[10] += 1
		when "l"
			alpha[11] += 1
		when "m"
			alpha[12] += 1
		when "n"
			alpha[13] += 1
		when "o"
			alpha[14] += 1
		when "p"
			alpha[15] += 1
		when "q"
			alpha[16] += 1
		when "r"
			alpha[17] += 1
		when "s"
			alpha[18] += 1
		when "t"
			alpha[19] += 1
		when "u"
			alpha[20] += 1
		when "v"
			alpha[21] += 1
		when "w"
			alpha[22] += 1
		when "x"
			alpha[23] += 1
		when "y"
			alpha[24] += 1
		when "z"
			alpha[25] += 1
	end
end

for s in 0..25
	print((s + 97).chr, "\s:\s", alpha[s], "\n")
end