Problem: When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he needs exactly m dishes. But at that, he doesn't want to order the same dish twice to taste as many dishes as possible. 

Kefa knows that the i-th dish gives him ai units of satisfaction. But some dishes do not go well together and some dishes go very well together. Kefa set to himself k rules of eating food of the following type — if he eats dish x exactly before dish y (there should be no other dishes between x and y), then his satisfaction level raises by c. 

Of course, our parrot wants to get some maximal possible satisfaction from going to the restaurant. Help him in this hard task!

Input

The first line of the input contains three space-separated numbers, n, m and k (1 ≤ m ≤ n ≤ 18, 0 ≤ k ≤ n * (n - 1)) — the number of dishes on the menu, the number of portions Kefa needs to eat to get full and the number of eating rules.

The second line contains n space-separated numbers ai, (0 ≤ ai ≤ 109) — the satisfaction he gets from the i-th dish.

Next k lines contain the rules. The i-th rule is described by the three numbers xi, yi and ci (1 ≤ xi, yi ≤ n, 0 ≤ ci ≤ 109). That means that if you eat dish xi right before dish yi, then the Kefa's satisfaction increases by ci. It is guaranteed that there are no such pairs of indexes i and j (1 ≤ i < j ≤ k), that xi = xj and yi = yj.

Output

In the single line of the output print the maximum satisfaction that Kefa can get from going to the restaurant.

Examples

Input

2 2 1
1 1
2 1 1


Output

3


Input

4 3 2
1 2 3 4
2 1 5
3 4 2


Output

12

Note

In the first sample it is best to first eat the second dish, then the first one. Then we get one unit of satisfaction for each dish and plus one more for the rule.

In the second test the fitting sequences of choice are 4 2 1 or 2 1 4. In both cases we get satisfaction 7 for dishes and also, if we fulfill rule 1, we get an additional satisfaction 5.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
bool IsPowerOfTwo(long long int x) { return (x != 0) && ((x & (x - 1)) == 0); }
long long int poer(long long int a, long long int b) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs *= a;
    a *= a;
    b /= 2;
  }
  return rs;
}
long long int mod_poer(long long int a, long long int b, long long int mo) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs = (rs * a) % mo;
    a = (a * a) % mo;
    b /= 2;
  }
  return rs % mo;
}
long long int d, x, y;
void ext_euclid(long long int a, long long int b) {
  if (b == 0) {
    d = a;
    x = 1;
    y = 0;
  } else {
    ext_euclid(b, a % b);
    long long int temp = x;
    x = y;
    y = temp - (a / b) * y;
  }
}
long long int mod_inverse1(long long int a, long long int m) {
  ext_euclid(a, m);
  return (x % m + m) % m;
}
long long int mod_inverse2(long long int a, long long int m) {
  return mod_poer(a, m - 2, m);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int fact[200005];
void fac() {
  fact[1] = fact[0] = 1;
  for (long long int i = 2; i <= 200005; ++i) {
    fact[i] = ((fact[i - 1] % MOD) * i) % MOD;
  }
}
long long int combi(long long int a, long long int b) {
  if (b > a) return 0;
  if (b == a)
    return 1;
  else {
    long long int x = mod_poer(fact[b], MOD - 2, MOD) % MOD;
    long long int y = mod_poer(fact[a - b], MOD - 2, MOD) % MOD;
    return ((((fact[a]) * x) % MOD) * y) % MOD;
  }
}
long long int n, m, k;
long long int dp[1 << 19][20];
long long int val[19];
map<pair<long long int, long long int>, long long int> mp;
long long int go(long long int mask, long long int pre, long long int co) {
  if (co == m) return 0;
  if (co > m) return -1e15;
  long long int ans = 0;
  if (dp[mask][pre] != -1) return dp[mask][pre];
  for (long long int i = 0; i < n; ++i) {
    if (!(mask & (1 << i))) {
      ans = max(ans,
                val[i] + mp[{pre + 1, i + 1}] + go(mask | (1 << i), i, co + 1));
    }
  }
  return dp[mask][pre] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (long long int i = 0; i < n; ++i) cin >> val[i];
  for (long long int i = 0; i < k; ++i) {
    long long int xi, yi, ci;
    cin >> xi >> yi >> ci;
    mp[{xi, yi}] = ci;
  }
  memset(dp, -1, sizeof(dp));
  cout << go(0, 19, 0) << endl;
}