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;
const int inf = 0x3f3f3f3f, mod = 1000000007;
const double pi = 3.1415926535897932, eps = 1e-6;
int n, k, m, f[102][102][27];
char s[105];
int c[27][27], ans = -inf;
int main() {
  scanf("%s%d%d", s, &k, &m);
  n = strlen(s);
  for (int(i) = (1); (i) <= (m); (i)++) {
    char t1[4], t2[4];
    int d;
    scanf("%s%s%d", t1, t2, &d);
    c[t1[0] - 97][t2[0] - 97] = d;
  }
  for (int(i) = (0); (i) <= (n); (i)++)
    for (int(j) = (0); (j) <= (k); (j)++)
      for (int(l) = (0); (l) <= (25); (l)++) f[i][j][l] = -inf;
  for (int(i) = (0); (i) <= (25); (i)++)
    if (i == s[0] - 97)
      f[0][0][i] = 0;
    else
      f[0][1][i] = 0;
  for (int(i) = (1); (i) < (n); (i)++)
    for (int(j) = (0); (j) <= (k); (j)++) {
      for (int(a) = (0); (a) <= (25); (a)++)
        for (int(b) = (0); (b) <= (25); (b)++) {
          if (b != s[i] - 97 && j == 0) continue;
          f[i][j][b] =
              max(f[i][j][b], f[i - 1][j - (b != s[i] - 97)][a] + c[a][b]);
        }
    }
  for (int(i) = (0); (i) <= (k); (i)++)
    for (int(j) = (0); (j) <= (25); (j)++) ans = max(ans, f[n - 1][i][j]);
  printf("%d", ans);
  return 0;
}
