Problem: You are given n strings ti. Each string has cost ci.

Let's define the function of string <image>, where ps, i is the number of occurrences of s in ti, |s| is the length of the string s. Find the maximal value of function f(s) over all strings.

Note that the string s is not necessarily some string from t.

Input

The first line contains the only integer n (1 ≤ n ≤ 105) — the number of strings in t.

Each of the next n lines contains contains a non-empty string ti. ti contains only lowercase English letters.

It is guaranteed that the sum of lengths of all strings in t is not greater than 5·105.

The last line contains n integers ci ( - 107 ≤ ci ≤ 107) — the cost of the i-th string.

Output

Print the only integer a — the maximal value of the function f(s) over all strings s. Note one more time that the string s is not necessarily from t.

Examples

Input

2
aa
bb
2 1


Output

4


Input

2
aa
ab
2 1


Output

5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Node {
  map<int, Node *> go;
  int l, r;
  Node *p, *link;
  Node(int _l, int _r, Node *_p) : l(_l), r(_r), p(_p), link(0) {}
};
const int MAXN = 100 * 1000 + 5, S = 1000;
string s[MAXN];
int c[MAXN], rr[MAXN], nn, curSuf;
long long ans = 0;
vector<int> a, sid;
Node *splitEdge(Node *v, int pos) {
  Node *res = new Node(v->l, pos, v->p);
  v->p->go[a[v->l]] = res;
  res->go[a[pos]] = v;
  v->p = res;
  v->l = pos;
  return res;
}
pair<Node *, int> go(Node *v, int l, int len) {
  while (len > 0) {
    v = v->go[a[l]];
    len -= v->r - v->l;
    l += v->r - v->l;
  }
  return make_pair(v, v->r + len);
}
void add(int i, pair<Node *, int> &p) {
  Node *w = 0;
  while (curSuf <= i) {
    Node *v = p.first;
    int pos = p.second;
    if (pos != v->r) {
      if (a[i] == a[pos]) {
        p.second++;
        return;
      }
      v = splitEdge(v, pos);
      Node *u = new Node(i, nn, v);
      v->go[a[i]] = u;
      if (w) w->link = v;
      w = v;
    } else {
      if (w) w->link = v;
      w = 0;
      if (v->go.find(a[i]) != v->go.end()) {
        p.first = v->go[a[i]];
        p.second = p.first->l + 1;
        return;
      }
      Node *u = new Node(i, nn, v);
      v->go[a[i]] = u;
    }
    int len = 0, l = v->r;
    Node *q = v;
    while (!q->link && q->p) {
      len += q->r - q->l;
      l -= q->r - q->l;
      q = q->p;
    }
    if (!q->p) {
      if (len) p = go(q, l + 1, len - 1);
    } else
      p = go(q->link, l, len);
    curSuf++;
  }
}
Node *buildSufTree() {
  nn = a.size();
  Node *root = new Node(0, 0, 0);
  pair<Node *, int> p = make_pair(root, 0);
  curSuf = 0;
  for (int i = 0; i < nn; i++) add(i, p);
  return root;
}
long long dfs(Node *v, int dep = 0) {
  long long res = 0;
  if (v->go.empty()) res += c[sid[nn - dep]];
  for (map<int, Node *>::iterator it = v->go.begin(); it != v->go.end(); it++)
    res += dfs(it->second, dep + it->second->r - it->second->l);
  if (v->go.empty()) {
    if (a[v->l] < S) ans = max(ans, res * (rr[sid[nn - dep]] - (nn - dep)));
  } else
    ans = max(ans, res * dep);
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  for (int i = 0; i < n; i++) {
    for (size_t j = 0; j < s[i].length(); j++) {
      a.push_back(s[i][j]);
      sid.push_back(i);
    }
    rr[i] = a.size();
    a.push_back(S + i);
    sid.push_back(i);
  }
  dfs(buildSufTree());
  cout << ans << '\n';
  return 0;
}