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;
void io() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
struct ACAM {
  int cnt, trie[200010][26], fail[200010], q[200010], sum[200010], h, r,
      num[200010];
  void insert(char s[], int n) {
    int now = 0;
    for (int i = 1; i <= n; ++i) {
      int p = s[i] - 'a';
      if (!trie[now][p]) trie[now][p] = ++cnt;
      now = trie[now][p];
    }
    sum[now]++;
  }
  void build() {
    h = 1;
    r = 0;
    for (int i = 0; i <= 25; ++i)
      if (trie[0][i]) q[++r] = trie[0][i];
    while (h <= r) {
      int u = q[h++];
      for (int i = 0; i <= 25; ++i) {
        if (trie[u][i]) {
          fail[trie[u][i]] = trie[fail[u]][i];
          sum[trie[u][i]] += sum[fail[trie[u][i]]];
          q[++r] = trie[u][i];
        } else
          trie[u][i] = trie[fail[u]][i];
      }
    }
  }
  void solve(char s[], int n) {
    int now = 0;
    for (int i = 1; i <= n; ++i) {
      int p = s[i] - 'a';
      now = trie[now][p];
      num[i] = sum[now];
    }
  }
} AC[2];
int n, len;
char s[200010], t[200010], rs[200010], rt[200010];
int main() {
  io();
  cin >> (t + 1) >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> (s + 1);
    len = strlen(s + 1);
    for (int i = len; i; --i) rs[len - i + 1] = s[i];
    AC[0].insert(s, len);
    AC[1].insert(rs, len);
  }
  len = strlen(t + 1);
  for (int i = len; i; --i) rt[len - i + 1] = t[i];
  AC[0].build();
  AC[1].build();
  AC[0].solve(t, len);
  AC[1].solve(rt, len);
  long long ans = 0;
  for (int i = 1; i <= len; ++i) ans += 1ll * AC[0].num[i] * AC[1].num[len - i];
  cout << ans;
}