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.
import java.util.*;
import java.io.*;

public class Solution {
    public void go() throws Exception {
        Scanner sc = new Scanner(new InputStreamReader(System.in));
        String s = sc.next();
        int k = sc.nextInt();
        int n = s.length();
        int cnt = sc.nextInt();
        int[][] add = new int[26][26];
        for (int i=0; i < cnt; i++) {
            String a = sc.next();
            String b = sc.next();
            add[a.charAt(0) - 'a'][b.charAt(0) - 'a'] = sc.nextInt();
        }
        sc.close();
                
        int[][][] f = new int[n][k + 1][26];
        for (int i=0; i < f.length; i++)
            for (int j=0; j < f[i].length; j++)
                Arrays.fill(f[i][j], -1000000000);
        
        for (char c = 'a'; c <= 'z'; c++)
            if (k > 0 || s.charAt(0) == c)
                f[0][c == s.charAt(0) ? 0 : 1][c - 'a'] = 0;
        
        for (int pos = 0; pos + 1 < n; pos++)
            for (int rep = 0; rep <= k; rep++)
                for (int last=0; last < 26; last++)
                    for (int next=0; next < 26; next++) {
                        int nextRep = rep + (s.charAt(pos + 1) - 'a' == next ? 0 : 1);
                        if (nextRep <= k)
                            f[pos + 1][nextRep][next] = Math.max(f[pos + 1][nextRep][next], f[pos][rep][last] + add[last][next]);
                    }
        
        int ans = -1000000000;
        for (int rep = 0; rep <= k; rep++)
            for (int last=0; last < 26; last++)
                ans = Math.max(ans, f[n-1][rep][last]);
        
        System.out.println(ans);
    }
    
    public static void main(String[] args) throws Exception {
        (new Solution()).go();
    }
}
