Problem: Barney was hanging out with Nora for a while and now he thinks he may have feelings for her. Barney wants to send her a cheesy text message and wants to make her as happy as possible.

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Initially, happiness level of Nora is 0. Nora loves some pickup lines like "I'm falling for you" and stuff. Totally, she knows n pickup lines, each consisting only of lowercase English letters, also some of them may be equal (in writing, but different in pronouncing or meaning though). Every time Nora sees i-th pickup line as a consecutive subsequence of Barney's text message her happiness level increases by ai. These substrings may overlap, for example, Nora will see the pickup line aa twice and the pickup line ab once in text message aaab.

Due to texting app limits, Barney's text may have up to l characters.

Barney asked you to help him make Nora as much happy as possible, it's gonna be legen...

Input

The first line of input contains two integers n and l (1 ≤ n ≤ 200, 1 ≤ l ≤ 1014) — the number of pickup lines and the maximum length of Barney's text.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100), meaning that Nora's happiness level increases by ai after every time seeing i-th pickup line.

The next n lines contain the pickup lines. i-th of them contains a single string si consisting of only English lowercase letter. Summary length of all pickup lines does not exceed 200.

All strings are not empty.

Output

Print the only integer — the maximum possible value of Nora's happiness level after reading Barney's text.

Examples

Input

3 6
3 2 1
heart
earth
art


Output

6


Input

3 6
3 2 8
heart
earth
art


Output

16

Note

An optimal answer for the first sample case is hearth containing each pickup line exactly once.

An optimal answer for the second sample case is artart.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool remin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool remax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
int nxt() {
  int first;
  scanf("%d", &first);
  return first;
}
long long gcd(long long a, long long b) {
  a = abs(a);
  b = abs(b);
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
struct pt {
  long double first, second;
  pt() {}
  pt(long double first, long double second) : first(first), second(second) {}
  inline pt operator-(const pt &r) const {
    return pt(first - r.first, second - r.second);
  }
  inline pt operator+(const pt &r) const {
    return pt(first + r.first, second + r.second);
  }
  inline pt operator*(const long double &r) const {
    return pt(first * r, second * r);
  }
  inline long double sqlen() const {
    return abs(first * first + second * second);
  }
  pt rot(long double co, long double si) const {
    return pt(first * co - si * second, si * first + co * second);
  }
  pt rot(long double ang) const { return rot(cosl(ang), sinl(ang)); }
  long double len() const { return sqrtl(sqlen()); }
  inline bool operator<(const pt &r) const {
    if (first != r.first) return first < r.first;
    return second < r.second;
  }
  inline bool operator==(const pt &r) const {
    return first == r.first && second == r.second;
  }
};
ostream &operator<<(ostream &os, const pt &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
inline long double cross(const pt &l, const pt &r) {
  return l.first * r.second - l.second * r.first;
}
inline long double dot(const pt &l, const pt &r) {
  return l.first * r.first + l.second * r.second;
}
long long pwmod(long long a, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
template <typename T>
inline T sqr(T first) {
  return first * first;
}
bool remin(long long &first, long long second) {
  if (first > second) {
    first = second;
    return 1;
  }
  return 0;
}
const int K = 26;
struct vertex {
  int next[K];
  vector<int> who;
  int p;
  char pch;
  int link;
  int go[K];
};
vertex t[400];
int sz;
void init() {
  t[0].p = t[0].link = -1;
  memset(t[0].next, 255, sizeof t[0].next);
  memset(t[0].go, 255, sizeof t[0].go);
  sz = 1;
}
void add_string(const string &s, int id) {
  int v = 0;
  for (size_t i = 0; i < s.length(); ++i) {
    char c = s[i] - 'a';
    if (t[v].next[c] == -1) {
      memset(t[sz].next, 255, sizeof t[sz].next);
      memset(t[sz].go, 255, sizeof t[sz].go);
      t[sz].link = -1;
      t[sz].p = v;
      t[sz].pch = c;
      t[v].next[c] = sz++;
    }
    v = t[v].next[c];
  }
  t[v].who.push_back(id);
}
int go(int v, char c);
int get_link(int v) {
  if (t[v].link == -1)
    if (v == 0 || t[v].p == 0)
      t[v].link = 0;
    else
      t[v].link = go(get_link(t[v].p), t[v].pch);
  return t[v].link;
}
int go(int v, char c) {
  if (t[v].go[c] == -1)
    if (t[v].next[c] != -1)
      t[v].go[c] = t[v].next[c];
    else
      t[v].go[c] = v == 0 ? 0 : go(get_link(v), c);
  return t[v].go[c];
}
long long cost[400][400];
const long long inf = LLONG_MAX / 2;
vector<vector<long long> > getAns(long long n) {
  if (n == 1) {
    vector<vector<long long> > ret(sz, vector<long long>(sz, 0));
    for (int i = 0; i < (int)(sz); i++) {
      for (int j = 0; j < (int)(sz); j++) {
        ret[i][j] = cost[i][j];
      }
    }
    return ret;
  }
  vector<vector<long long> > cur(sz, vector<long long>(sz, -inf));
  vector<vector<long long> > ch = getAns(n / 2);
  for (int k = 0; k < (int)(sz); k++) {
    for (int i = 0; i < (int)(sz); i++) {
      for (int j = 0; j < (int)(sz); j++) {
        remax(cur[i][j], ch[i][k] + ch[k][j]);
      }
    }
  }
  for (int i = 0; i < (int)(sz); i++) {
    for (int j = 0; j < (int)(sz); j++) {
    }
  }
  vector<vector<long long> > ret;
  if (n & 1) {
    ret.assign(sz, vector<long long>(sz, -inf));
    for (int k = 0; k < (int)(sz); k++) {
      for (int i = 0; i < (int)(sz); i++) {
        for (int j = 0; j < (int)(sz); j++) {
          remax(ret[i][j], cur[i][k] + cost[k][j]);
        }
      }
    }
  } else {
    ret.swap(cur);
  }
  return ret;
}
void solve() {
  int n;
  long long l;
  cin >> n >> l;
  vector<int> a(n);
  for (int i = 0; i < (int)(n); i++) {
    cin >> a[i];
  }
  vector<string> s(n);
  for (int i = 0; i < (int)(n); i++) {
    cin >> s[i];
  }
  init();
  for (int i = 0; i < (int)(n); i++) {
    add_string(s[i], i);
  }
  long long add_cost[sz];
  memset(add_cost, 0, sizeof(add_cost));
  for (int i = 0; i < (int)(sz); i++) {
    int v = i;
    while (v != 0) {
      for (int id : t[v].who) {
        add_cost[i] += a[id];
      }
      v = get_link(v);
    }
  }
  for (int i = 0; i < (int)(sz); i++) {
    for (int j = 0; j < (int)(sz); j++) {
      cost[i][j] = -inf;
    }
  }
  for (int i = 0; i < (int)(sz); i++) {
    for (int j = 0; j < (int)(26); j++) {
      int v = go(i, j);
      remax(cost[i][v], add_cost[v]);
    }
  }
  auto ans = getAns(l);
  long long ret = 0;
  for (int i = 0; i < (int)(sz); i++) {
    remax(ret, ans[0][i]);
  }
  cout << ret << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}