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>
const int maxn = 2e5 + 5;
int n, lent;
char s[maxn], t[maxn];
struct ACAM {
  int ch[maxn][26], tag[maxn], fail[maxn], cnt[maxn], tot = 1;
  int match[maxn];
  void insert(char *s) {
    int now = 1, len = strlen(s);
    for (int i = 0; i < len; ++i) {
      int sid = s[i] - 'a';
      if (!ch[now][sid]) ch[now][sid] = ++tot;
      now = ch[now][sid];
    }
    ++tag[now];
  }
  void build_fail() {
    std ::queue<int> q;
    fail[1] = 1;
    for (int i = 0; i < 26; ++i)
      if (ch[1][i])
        fail[ch[1][i]] = 1, q.push(ch[1][i]);
      else
        ch[1][i] = 1;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      int f = fail[u];
      for (int i = 0; i < 26; ++i) {
        int &v = ch[u][i];
        if (v)
          fail[v] = ch[f][i], q.push(v);
        else
          v = ch[f][i];
      }
      cnt[u] = tag[u] + cnt[fail[u]];
    }
  }
  void Match(char *t) {
    int now = 1, len = strlen(t);
    for (int i = 0; i < len; ++i) {
      now = ch[now][t[i] - 'a'];
      match[i] = cnt[now];
    }
  }
} F, G;
int main() {
  scanf("%s", t);
  lent = strlen(t);
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s), F.insert(s);
    std ::reverse(s, s + strlen(s));
    G.insert(s);
  }
  F.build_fail(), G.build_fail();
  F.Match(t);
  std ::reverse(t, t + lent);
  G.Match(t);
  long long ans = 0;
  for (int i = 0; i < lent - 1; ++i)
    ans = ans + (long long)F.match[i] * G.match[lent - i - 2];
  printf("%lld", ans);
  return 0;
}