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: Accepted)

def get_i() #空白区切の入力を数値(整数)の配列で返す
  return gets.chomp.split(" ").map(&:to_i)
end
def get_f() #空白区切の入力を数値(実数)の配列で返す
  return gets.chomp.split(" ").map(&:to_f)
end
def get() #空白区切の入力を文字列の配列で返す
  return gets.chomp.split(" ")
end
def get_nsp() #入力されたものを一文字ずつに区切った文字列の配列で返す
  return gets.chomp.split("")
end
def yn_judge(bool,y="Yes",n="No") #boolに真偽を投げることで、trueならy、falseならnの値を出力する
  return bool ? y : n 
end
def array(size,n=1,init=nil) #nに配列の次元数、sizeに配列の大きさ、initに初期値を投げることでその配列を返す
  if n==1
    return Array.new(size){init}
  else
    return Array.new(n).map{Array.new(size){init}}
  end
end

s=""
while true
  s_input=gets
  break if s_input.nil?
  break if s_input.chomp==""
  s_input=s_input.chomp.split("")
  s+=s_input.join("").downcase
  #break if s_input[s_input.size-1]=="."
end
alphabet=("a".."z").to_a
26.times do|i|
  puts "#{alphabet[i]} : #{s.count(alphabet[i])}"
end