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 N = 1e6 + 5;
struct AC_auto {
  int tr[N][26]{}, tot{};
  int val[N]{}, fail[N]{};
  int dp1[N]{}, sum1[N]{};
  vector<int> v1;
  void insert(const char *s) {
    int u = 0, len1 = strlen(s);
    for (int i = 0; i < len1; ++i) {
      if (!tr[u][s[i] - 'a']) tr[u][s[i] - 'a'] = ++tot;
      u = tr[u][s[i] - 'a'];
    }
    val[u]++;
  }
  void build() {
    queue<int> q;
    for (int i = 0; i < 26; ++i)
      if (tr[0][i]) q.push(tr[0][i]);
    while (!q.empty()) {
      int u = q.front();
      v1.push_back(u);
      q.pop();
      for (int i = 0; i < 26; ++i) {
        if (tr[u][i])
          fail[tr[u][i]] = tr[fail[u]][i], q.push(tr[u][i]);
        else
          tr[u][i] = tr[fail[u]][i];
      }
    }
    for (int u : v1) {
      dp1[u] = dp1[fail[u]] + val[u];
    }
  }
  void query(const char *t) {
    int u = 0;
    for (int i = 0; t[i]; ++i) {
      u = tr[u][t[i] - 'a'];
      sum1[i] = dp1[u];
    }
  }
} ac1, ac2;
char t[N], s1[N];
int main() {
  scanf("%s", t);
  int n = 0;
  scanf("%d", &n);
  for (int ccc = 0; ccc < n; ++ccc) {
    scanf("%s", s1);
    ac1.insert(s1);
    reverse(s1, s1 + strlen(s1));
    ac2.insert(s1);
  }
  ac1.build();
  ac2.build();
  long long ans1 = 0;
  ac1.query(t);
  reverse(t, t + strlen(t));
  ac2.query(t);
  int len1 = strlen(t);
  for (int i = 1; i <= len1; ++i)
    ans1 += (long long)ac1.sum1[i - 1] * ac2.sum1[len1 - i - 1];
  printf("%lld\n", ans1);
  return 0;
}