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 = 2e5 + 10;
int f1[N], f2[N];
struct node {
  int tr[N][30], fail[N], tot;
  int w[N];
  void inse(char ss[], int len) {
    int u = 1;
    for (int i = 1; i <= len; i++) {
      int ch = ss[i] - 'a';
      if (!tr[u][ch]) tr[u][ch] = ++tot;
      u = tr[u][ch];
    }
    w[u]++;
  }
  void init() {
    for (int i = 0; i < 26; i++) tr[0][i] = 1;
    queue<int> q;
    q.push(1);
    while (!q.empty()) {
      int p = q.front();
      q.pop();
      for (int i = 0; i < 26; i++) {
        if (tr[p][i])
          fail[tr[p][i]] = tr[fail[p]][i], w[tr[p][i]] += w[fail[tr[p][i]]],
          q.push(tr[p][i]);
        else
          tr[p][i] = tr[fail[p]][i];
      }
    }
  }
  void cha(char ss[], int len, int f[]) {
    int u = 1;
    for (int i = 1; i <= len; i++) {
      u = tr[u][ss[i] - 'a'];
      f[i] = w[u];
    }
  }
} l1, l2;
int n, lent, m;
char t[N];
char s[N];
int main() {
  scanf("%s", t + 1);
  lent = strlen(t + 1);
  cin >> n;
  l1.tot = l2.tot = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    m = strlen(s + 1);
    l1.inse(s, m);
    reverse(s + 1, s + m + 1);
    l2.inse(s, m);
  }
  l1.init(), l2.init();
  l1.cha(t, lent, f1);
  reverse(t + 1, t + lent + 1);
  l2.cha(t, lent, f2);
  long long ans = 0;
  for (int i = 1; i < lent; i++) ans += 1ll * f1[i] * f2[lent - i];
  cout << ans;
}