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[200010], t[200010];
queue<int> q;
struct AC_Automation {
  int ch[200010][30], cnt[200010], tot;
  int fail[200010];
  void insert(char *s, int len) {
    int now = 0;
    for (int i = 1; i <= len; i++) {
      int cur = s[i] - 'a';
      if (!ch[now][cur]) ch[now][cur] = ++tot;
      now = ch[now][cur];
    }
    cnt[now]++;
  }
  void get_fail() {
    for (int i = 0; i < 26; i++)
      if (ch[0][i]) q.push(ch[0][i]);
    while (q.size()) {
      int now = q.front();
      q.pop();
      cnt[now] += cnt[fail[now]];
      for (int i = 0; i < 26; i++) {
        if (!ch[now][i])
          ch[now][i] = ch[fail[now]][i];
        else {
          fail[ch[now][i]] = ch[fail[now]][i];
          q.push(ch[now][i]);
        }
      }
    }
  }
} A, B;
int f[200010], g[200010];
int main() {
  scanf("%s", t + 1);
  n = strlen(t + 1);
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    A.insert(s, len);
    reverse(s + 1, s + len + 1);
    B.insert(s, len);
  }
  A.get_fail();
  B.get_fail();
  int now = 0;
  for (int i = 1; i <= n; i++) {
    now = A.ch[now][t[i] - 'a'];
    f[i] = A.cnt[now];
  }
  now = 0;
  for (int i = n; i >= 1; i--) {
    now = B.ch[now][t[i] - 'a'];
    g[i] = B.cnt[now];
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += 1ll * f[i] * g[i + 1];
  printf("%lld\n", ans);
  return 0;
}