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;
#pragma comment(linker, "/STACK:64000000")
template <class T>
inline T sqr(T x) {
  return x * x;
}
int n;
string s;
int cost[26][26] = {0};
int k;
char used[128][128][26] = {0};
int d[128][128][26];
int rec(int x, int left, int last) {
  if (x == n) return 0;
  int &res = d[x][left][last];
  if (used[x][left][last]) return res;
  used[x][left][last] = true;
  res = rec(x + 1, left, s[x]) + (x ? cost[last][s[x]] : 0);
  for (int i = 0; left && i < 26; ++i)
    if (i != s[x])
      res = max(res, rec(x + 1, left - 1, i) + (x ? cost[last][i] : 0));
  return res;
}
int main() {
  cin >> s >> k;
  n = (int)((s).size());
  for (int i = 0; i < n; ++i) s[i] -= 'a';
  int m;
  cin >> m;
  int z;
  for (char x, y; m--;) {
    cin >> x >> y >> z;
    x -= 'a', y -= 'a';
    cost[x][y] = z;
  }
  cout << rec(0, k, 0) << endl;
  return 0;
}
