Problem: You are given a string t and n strings s_1, s_2, ..., s_n. All strings consist of lowercase Latin letters.

Let f(t, s) be the number of occurences of string s in string t. For example, f('aaabacaa', 'aa') = 3, and f('ababa', 'aba') = 2.

Calculate the value of ∑_{i=1}^{n} ∑_{j=1}^{n} f(t, s_i + s_j), where s + t is the concatenation of strings s and t. Note that if there are two pairs i_1, j_1 and i_2, j_2 such that s_{i_1} + s_{j_1} = s_{i_2} + s_{j_2}, you should include both f(t, s_{i_1} + s_{j_1}) and f(t, s_{i_2} + s_{j_2}) in answer.

Input

The first line contains string t (1 ≤ |t| ≤ 2 ⋅ 10^5).

The second line contains integer n (1 ≤ n ≤ 2 ⋅ 10^5).

Each of next n lines contains string s_i (1 ≤ |s_i| ≤ 2 ⋅ 10^5).

It is guaranteed that ∑_{i=1}^{n} |s_i| ≤ 2 ⋅ 10^5. All strings consist of lowercase English letters.

Output

Print one integer — the value of ∑_{i=1}^{n} ∑_{j=1}^{n} f(t, s_i + s_j).

Examples

Input


aaabacaa
2
a
aa


Output


5


Input


aaabacaa
4
a
a
a
b


Output


33

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct AC {
  int cnt = 0;
  AC *ch[26]{}, *fail;
  AC() {}
};
AC *rt = new AC();
void bt(string t) {
  AC *now = rt;
  for (char i : t) {
    if (!now->ch[i - 'a']) now->ch[i - 'a'] = new AC();
    now = now->ch[i - 'a'];
  }
  now->cnt++;
}
void bf() {
  queue<AC *> q;
  for (AC *i : rt->ch)
    if (i) q.push(i), i->fail = rt;
  while (!q.empty()) {
    auto p = q.front();
    q.pop();
    for (int i = 0; i < 26; i++) {
      if (!p->ch[i]) continue;
      AC *ptr = p->fail;
      while (ptr != rt and !ptr->ch[i]) ptr = ptr->fail;
      if (ptr->ch[i]) ptr = ptr->ch[i];
      p->ch[i]->fail = ptr;
      p->ch[i]->cnt += ptr->cnt;
      q.push(p->ch[i]);
    }
  }
}
vector<int> match(string s) {
  AC *ptr = rt;
  vector<int> ans;
  ans.assign(s.size(), 0);
  for (int i = 0; i < s.size(); i++) {
    while (ptr != rt and !ptr->ch[s[i] - 'a']) ptr = ptr->fail;
    if (ptr->ch[s[i] - 'a']) ptr = ptr->ch[s[i] - 'a'];
    ans[i] = ptr->cnt;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  int m;
  cin >> m;
  string t[m];
  for (string &i : t) cin >> i, bt(i);
  bf();
  vector<int> pre = match(s);
  rt = new AC();
  for (string i : t) {
    reverse(i.begin(), i.end());
    bt(i);
  }
  reverse(s.begin(), s.end());
  bf();
  vector<int> suf = match(s);
  reverse(suf.begin(), suf.end());
  long long ans = 0;
  for (int i = 0; i < s.size() - 1; i++) ans += 1ll * pre[i] * suf[i + 1];
  cout << ans;
}