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;
const int Sz = 2e5 + 5;
int f1[Sz], f2[Sz], n, m;
char s[Sz], t[Sz];
struct AC {
  int ch[Sz][26], bo[Sz], tot = 1, nxt[Sz];
  int val[Sz];
  inline void ins(char *s) {
    int u = 1;
    for (int i = 0, len = strlen(s); i < len; i++) {
      int c = s[i] - 'a', &v = ch[u][c];
      if (!v) v = ++tot;
      u = v;
    }
    val[u]++;
  }
  inline void bfs() {
    for (int i = 0; i < 26; i++) ch[0][i] = 1;
    queue<int> q;
    q.push(1);
    nxt[1] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int i = 0; i < 26; i++) {
        int &v = ch[u][i];
        if (!v)
          v = ch[nxt[u]][i];
        else
          q.push(v), nxt[v] = ch[nxt[u]][i], val[v] += val[nxt[v]];
      }
    }
  }
  inline void query(char *s, int *f) {
    for (int i = 0, u = 1, len = strlen(s); i < len; i++) {
      u = ch[u][s[i] - 'a'];
      f[i] = val[u];
    }
  }
} A, B;
int main() {
  scanf("%s", t);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    A.ins(s);
    std::reverse(s, s + strlen(s));
    B.ins(s);
  }
  A.bfs();
  B.bfs();
  A.query(t, f1);
  std::reverse(t, t + strlen(t));
  B.query(t, f2);
  std::reverse(f2, f2 + strlen(t));
  long long ans = 0;
  for (int i = 0, len = strlen(t); i < len - 1; i++)
    ans += 1ll * f1[i] * f2[i + 1];
  cout << ans;
}