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.
#include <bits/stdc++.h>
using namespace std;
int const MAX_SIZE = 110;
int const MAX_K = 110;
int dp[MAX_SIZE][MAX_K][26];
int main() {
  string word;
  int max_k;
  cin >> word >> max_k;
  int n;
  cin >> n;
  map<pair<int, int>, int> t;
  for (int i = (0); i < ((n)); ++i) {
    char a, b;
    int c;
    cin >> a >> b >> c;
    t[make_pair(a - 'a', b - 'a')] = c;
  }
  int const INF = 1e9;
  for (int i = (0); i < ((MAX_SIZE)); ++i)
    for (int j = (0); j < ((MAX_K)); ++j)
      for (int k = (0); k < ((26)); ++k) dp[i][j][k] = -INF;
  for (int i = (0); i < ((26)); ++i) dp[1][i != word[0] - 'a'][i] = 0;
  for (int i = (1); i < ((int)word.size()); ++i)
    for (int j = (0); j < ((max_k + 1)); ++j)
      for (int k1 = (0); k1 < ((26)); ++k1)
        for (int k2 = (0); k2 < ((26)); ++k2)
          if (dp[i][j][k1] != -INF) {
            int cost = 0;
            if (t.count(make_pair(k1, k2))) cost = t[make_pair(k1, k2)];
            int& next = dp[i + 1][j + (k2 != word[i] - 'a')][k2];
            if (next < dp[i][j][k1] + cost) {
              next = dp[i][j][k1] + cost;
            }
          }
  int result = -INF;
  for (int i = (0); i < ((max_k + 1)); ++i)
    for (int j = (0); j < ((26)); ++j)
      result = max(result, dp[(int)word.size()][i][j]);
  cout << result << endl;
  return 0;
}
