Vasya plays the LionAge II. He was bored of playing with a stupid computer, so he installed this popular MMORPG, to fight with his friends. Vasya came up with the name of his character — non-empty string s, consisting of a lowercase Latin letters. However, in order not to put up a front of friends, Vasya has decided to change no more than k letters of the character name so that the new name sounded as good as possible. Euphony of the line is defined as follows: for each pair of adjacent letters x and y (x immediately precedes y) the bonus c(x, y) is added to the result. Your task is to determine what the greatest Euphony can be obtained by changing at most k letters in the name of the Vasya's character.

Input

The first line contains character's name s and an integer number k (0 ≤ k ≤ 100). The length of the nonempty string s does not exceed 100. The second line contains an integer number n (0 ≤ n ≤ 676) — amount of pairs of letters, giving bonus to the euphony. The next n lines contain description of these pairs «x y c», which means that sequence xy gives bonus c (x, y — lowercase Latin letters,  - 1000 ≤ c ≤ 1000). It is guaranteed that no pair x y mentioned twice in the input data.

Output

Output the only number — maximum possible euphony оf the new character's name.

Examples

Input

winner 4
4
s e 7
o s 8
l o 13
o o 8


Output

36

Input

abcdef 1
5
a b -10
b c 5
c d 5
d e 5
e f 5


Output

20

Note

In the first example the most euphony name will be looser. It is easy to calculate that its euphony is 36.
//package round66;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {

	static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter out = new PrintWriter(System.out);
	
	static String nextToken() throws IOException{
		while (st==null || !st.hasMoreTokens()){
			String s = bf.readLine();
			if (s == null)
				return null;
			st = new StringTokenizer(s);
		}
		
		return st.nextToken();
	}
	
	static int nextInt() throws IOException{
		return Integer.parseInt(nextToken());
	}
	
	static long nextLong() throws IOException{
		return Long.parseLong(nextToken());
	}
	
	static String nextStr() throws IOException{
		return nextToken();
	}
	
	static double nextDouble() throws IOException{
		return Double.parseDouble(nextToken());
	}
	
	public static void main(String[] args) throws IOException{
		byte s[] = nextStr().getBytes();
		int k = nextInt(),
			n = nextInt(),
			l = s.length;
		k = Math.min(k, l);
		
		for (int i=0; i<l; i++)
			s[i] -= 'a';
		
		int c[][] = new int[26][26];
		for (int i=0; i<n; i++){
			byte c1 = (byte) (nextStr().charAt(0)-'a');
			byte c2 = (byte) (nextStr().charAt(0)-'a');
			int p = nextInt();
			c[c1][c2] = p;
		}
		
		int inf = -1000000000;
		int a[][][] = new int[l][k+1][26];
		for (int i=0; i<26; i++)
			a[0][0][i] = inf;
		a[0][0][s[0]] = 0;
		
		for (int i=1; i<l; i++){
			for (int p=0; p<26; p++)
				a[i][0][p] = inf;
			a[i][0][s[i]] = a[i-1][0][s[i-1]]+c[s[i-1]][s[i]];
			
			for (int j=1; j<=k; j++){
				for (int p=0; p<26; p++)
					a[i][j][p] = inf;
				for (int p=0; p<26; p++)
					a[i][j][s[i]] = Math.max(a[i][j][s[i]], a[i-1][j][p]+c[p][s[i]]);
				
				
				for (int t=0; t<26; t++){
					for (int p=0; p<26; p++){
						if (j>1 || p==s[i-1])
							a[i][j][t] = Math.max(a[i][j][t], a[i-1][j-1][p]+c[p][t]);
					}
				}
			}
		}
		
		int res = inf;
		for (int i=0; i<=k; i++)
			for (int j=0; j<26; j++)
				res = Math.max(res, a[l-1][i][j]);
		
		out.println(res);
		out.flush();

	}

}
