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 maxn = 101010 * 4;
struct ACMachine {
  int tr[maxn][26];
  int fail[maxn];
  int nume;
  int cnt[maxn];
  vector<int> vec[maxn];
  queue<int> q;
  void init() {
    while (!q.empty()) q.pop();
    nume = 0;
  }
  int addnode() {
    ++nume;
    for (int i = 0; i < 26; i++) {
      tr[nume][i] = 0;
    }
    cnt[nume] = fail[nume] = 0;
    return nume;
  }
  int insert(int x, int c) {
    c -= 'a';
    if (!tr[x][c]) {
      tr[x][c] = addnode();
    }
    return tr[x][c];
  }
  void build() {
    for (int i = 0; i < 26; i++)
      if (tr[0][i]) {
        q.push(tr[0][i]);
        vec[0].push_back(tr[0][i]);
      }
    while (q.size()) {
      int u = q.front();
      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]);
          vec[fail[tr[u][i]]].push_back(tr[u][i]);
        } else
          tr[u][i] = tr[fail[u]][i];
      }
    }
  }
  void dfs(int x, int val) {
    int siz = vec[x].size();
    cnt[x] += val;
    for (int i = 0; i < siz; i++) {
      dfs(vec[x][i], cnt[x]);
    }
  }
} ac1, ac2;
char ch[maxn];
char temv[maxn];
int main() {
  int n;
  scanf("%s%d", ch, &n);
  int pos1 = 0, pos2 = 0, len = strlen(ch);
  ac1.init();
  ac2.init();
  for (int j = 0; j < len; j++) pos1 = ac1.insert(pos1, ch[j]);
  for (int j = len - 1; j >= 0; j--) pos2 = ac2.insert(pos2, ch[j]);
  for (int i = 1; i <= n; i++) {
    scanf("%s", temv);
    len = strlen(temv);
    pos1 = 0;
    pos2 = 0;
    for (int j = 0; j < len; j++) pos1 = ac1.insert(pos1, temv[j]);
    for (int j = len - 1; j >= 0; j--) pos2 = ac2.insert(pos2, temv[j]);
    ac1.cnt[pos1]++;
    ac2.cnt[pos2]++;
  }
  len = strlen(ch);
  ac1.build();
  ac2.build();
  ac1.dfs(0, 0);
  ac2.dfs(0, 0);
  len = strlen(ch);
  long long ans = 0;
  for (int i = 1; i < len; i++) {
    ans += (long long)ac1.cnt[i] * ac2.cnt[len - i];
  }
  printf("%lld\n", ans);
}