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 C (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>   
 

int main()
{
int count=0,i,j;
 char str[1200];
 char array[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
 int dem[26]={0};
while(1){
        scanf("%s",&str); 
        if (str[0]=='0')break;
        for (i = 0; i < 26; i++)
		{
			for(j=0;str[j]!='\0';j++){
				if(array[i]==str[j])
				{
					dem[i]=dem[i] +1;
				}
			}
			printf("%c:%d\n",array[i],dem[i]);
         }
        }
 return 0;
}