Fibonacci String

After little Jim learned Fibonacci Number in the class , he was very interest in it.
Now he is thinking about a new thing -- Fibonacci String .

He defines : str[n] = str[n-1] + str[n-2] ( n > 1 )

He is so crazying that if someone gives him two strings str[0] and str[1], he will calculate the str[2],str[3],str[4] , str[5]....

For example :
If str[0] = "ab"; str[1] = "bc";
he will get the result , str[2]="abbc", str[3]="bcabbc" , str[4]="abbcbcabbc" …………;

As the string is too long ,Jim can't write down all the strings in paper. So he just want to know how many times each letter appears in Kth Fibonacci String . Can you help him ?

Input

The first line contains a integer N which indicates the number of test cases.
Then N cases follow.
In each case,there are two strings str[0], str[1] and a integer K (0 <= K < 50) which are separated by a blank.
The string in the input will only contains less than 30 low-case letters.

Output

For each case,you should count how many times each letter appears in the Kth Fibonacci String and print out them in the format "X:N".
If you still have some questions, look the sample output carefully.
Please output a blank line after each test case.

To make the problem easier, you can assume the result will in the range of int.

Sample Input

1
ab bc 3

Sample Output

a:1
b:3
c:2
d:0
e:0
f:0
g:0
h:0
i:0
j:0
k:0
l:0
m:0
n:0
o:0
p:0
q:0
r:0
s:0
t:0
u:0
v:0
w:0
x:0
y:0
z:0

#include<iostream>
#include<string>
#include<string.h>
using namespace std;
int t,k;

struct Num{
	int z[30];
}n[60];
int main()
{
	cin>>t;
	while(t--)
	{	
		string a,b;
		cin>>a>>b>>k;
		for(int i=0;i<=k;i++)
		memset(n[i].z,0,sizeof(n[i].z));
		for(int i=0;i<a.size();i++)
		{
			int j=a[i]-'a';
			n[0].z[j]++;
			
		}
		
		for(int i=0;i<b.size();i++)
		{
			int j=b[i]-'a';
			n[1].z[j]++;
		}
		for(int j=2;j<=k;j++){
			for(int i=0;i<30;i++)
			{
				n[j].z[i]=n[j-1].z[i]+n[j-2].z[i];	
				//cout<<n[j].z[i];
			}
			
		}
		char c='a';
		for(int i=0;i<26;i++)
		{
			cout<<c<<':'<<n[k].z[i]<<endl;
			c++;
		}
		cout<<endl;
	}
	return 0;
}