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 JAVA (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int[] alpha = new int[27];//0-26
		for (int syoki = 0; syoki < 27; syoki++) {
			alpha[syoki] = 0;
		}

		/*文字と数値一覧表
		 * A===10===a
		 * B===11===b
		 * !
		 * !
		 * !
		 * Z===35==z
		 * alpha[0]=a--->Character.getNumericValue('A')-10;
		 * alpha[1]=b--->Character.getNumericValue('B')-10;
		 */

		//個数のカウント
		for (; sc.hasNextLine();) {
			int sum = 0;
			char[] swap = sc.nextLine().toCharArray();//char型の配列に格納する

			for (int i = 0; i < swap.length; i++) {
				if (swap[i] != ' ') {
					int a = Character.getNumericValue(swap[i]);
					//  System.out.println("aは"+a);
					//  System.out.println("a:" + a);
					if (a >= 10 && a < 37) {
						alpha[a - 10]++;
					} else {
						//break;
					}
				}
			}
		}
		//A~Zの個数を出力
		char L00 = 'a';
		for (int L01 = 0, i = 0; L01 <= ('z' - 'a'); L01++, i++) {
			System.out.print(L00++);
			System.out.println(" : " + alpha[i]);
		}

	}

}