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;
struct Trie {
  struct node {
    int fail, cnt, ch[26];
  } T[200010];
  int w[200010];
  int tot, root;
  void init() {
    tot = root = 0;
    memset(T, 0, sizeof(T));
  }
  void ins(char* x) {
    int o = root, len = strlen(x);
    for (int k = 0; k < len; k++) {
      int c = x[k] - 'a';
      if (!T[o].ch[c]) T[o].ch[c] = ++tot;
      o = T[o].ch[c];
    }
    T[o].cnt++;
  }
  void get_fail() {
    queue<int> q;
    for (int i = 0; i < 26; i++)
      if (T[0].ch[i]) q.push(T[0].ch[i]);
    while (!q.empty()) {
      int o = q.front();
      q.pop();
      for (int i = 0; i < 26; i++) {
        if (T[o].ch[i]) {
          T[T[o].ch[i]].fail = T[T[o].fail].ch[i];
          T[T[o].ch[i]].cnt += T[T[T[o].ch[i]].fail].cnt;
          q.push(T[o].ch[i]);
        } else
          T[o].ch[i] = T[T[o].fail].ch[i];
      }
    }
  }
  void AC_automation(char* x) {
    int o = root, len = strlen(x);
    for (int i = 0; i < len; i++) {
      int j = o = T[o].ch[x[i] - 'a'];
      if (T[j].cnt > 0) w[i] += T[j].cnt;
      j = T[j].fail;
    }
  }
} T1, T2;
char t[200010], s[200010];
int main() {
  scanf("%s", t);
  int n;
  scanf("%d", &n);
  while (n--) {
    scanf("%s", s);
    T1.ins(s);
    int l = strlen(s);
    reverse(s, s + l);
    T2.ins(s);
  }
  T1.get_fail();
  T2.get_fail();
  T1.AC_automation(t);
  int l = strlen(t);
  reverse(t, t + l);
  T2.AC_automation(t);
  long long ans = 0;
  for (int i = 0; i < l - 1; i++) ans += 1ll * T1.w[i] * T2.w[l - i - 2];
  cout << ans;
}