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;
int n;
char s[200100], t[200100];
struct ACM {
  int tr[200100][26], val[200100], cnt = 1, fail[200100];
  void insert(char t[]) {
    int now = 0;
    for (int i = 0; t[i]; ++i) {
      int &nxt = tr[now][t[i] - 'a'];
      if (!nxt) nxt = cnt++;
      now = nxt;
    }
    val[now]++;
  }
  void getfail() {
    queue<int> q;
    for (int i = 0; i < 26; ++i) {
      if (tr[0][i]) q.push(tr[0][i]);
    }
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      val[now] += val[fail[now]];
      for (int i = 0; i < 26; ++i) {
        if (!tr[now][i]) {
          tr[now][i] = tr[fail[now]][i];
        } else {
          fail[tr[now][i]] = tr[fail[now]][i];
          q.push(tr[now][i]);
        }
      }
    }
  }
} pre, suf;
int a[200100], b[200100];
int main() {
  scanf("%s%d", s, &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", &t);
    pre.insert(t);
    reverse(t, t + strlen(t));
    suf.insert(t);
  }
  pre.getfail();
  suf.getfail();
  int now = 0;
  for (int i = 0; s[i]; ++i) {
    now = pre.tr[now][s[i] - 'a'];
    a[i] = pre.val[now];
  }
  reverse(s, s + strlen(s));
  now = 0;
  for (int i = 0; s[i]; ++i) {
    now = suf.tr[now][s[i] - 'a'];
    b[i] = suf.val[now];
  }
  reverse(b, b + strlen(s));
  long long ans = 0;
  for (int i = 1; s[i]; ++i) {
    ans += 1LL * a[i - 1] * b[i];
  }
  cout << ans << endl;
  return 0;
}