#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
struct AC {
  int tr[N][26], cnt;
  int fail[N];
  int ans[N];
  int exist[N];
  int idx[N];
  void cle() {
    cnt = 0;
    memset(tr, 0, sizeof tr);
    memset(fail, 0, sizeof fail);
    memset(ans, 0, sizeof ans);
    memset(idx, 0, sizeof idx);
    memset(exist, 0, sizeof exist);
  }
  void insert(char* s) {
    int u = 0;
    for (int i = 1; s[i]; i++) {
      if (!tr[u][s[i] - 'a']) tr[u][s[i] - 'a'] = ++cnt;
      u = tr[u][s[i] - 'a'];
    }
    exist[u]++;
  }
  queue<int> q;
  void build() {
    for (int i = 0; i < 26; i++)
      if (tr[0][i]) q.push(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],
          exist[tr[u][i]] += exist[fail[tr[u][i]]], q.push(tr[u][i]);
        else
          tr[u][i] = tr[fail[u]][i];
      }
    }
  }
  void query(char* t) {
    int u = 0;
    for (int i = 1; t[i]; i++) {
      u = tr[u][t[i] - 'a'];
      ans[i] = exist[u];
    }
  }
} T1, T2;
char s[N];
char t[2000006];
int n;
int main() {
  scanf("%s", t + 1);
  int n;
  cin >> n;
  while (n--) {
    scanf("%s", s + 1);
    T1.insert(s);
    int l = strlen(s + 1);
    for (int i = 1; i <= l / 2; i++) swap(s[i], s[l - i + 1]);
    T2.insert(s);
  }
  T1.build();
  T2.build();
  T1.query(t);
  int l = strlen(t + 1);
  for (int i = 1; i <= l / 2; i++) swap(t[i], t[l - i + 1]);
  T2.query(t);
  long long res = 0;
  for (int i = 1; i <= l; i++) {
    res += 1ll * T1.ans[i] * T2.ans[l - i];
  }
  cout << res << ' ';
}

### Prompt

Generate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
struct AC {
  int tr[N][26], cnt;
  int fail[N];
  int ans[N];
  int exist[N];
  int idx[N];
  void cle() {
    cnt = 0;
    memset(tr, 0, sizeof tr);
    memset(fail, 0, sizeof fail);
    memset(ans, 0, sizeof ans);
    memset(idx, 0, sizeof idx);
    memset(exist, 0, sizeof exist);
  }
  void insert(char* s) {
    int u = 0;
    for (int i = 1; s[i]; i++) {
      if (!tr[u][s[i] - 'a']) tr[u][s[i] - 'a'] = ++cnt;
      u = tr[u][s[i] - 'a'];
    }
    exist[u]++;
  }
  queue<int> q;
  void build() {
    for (int i = 0; i < 26; i++)
      if (tr[0][i]) q.push(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],
          exist[tr[u][i]] += exist[fail[tr[u][i]]], q.push(tr[u][i]);
        else
          tr[u][i] = tr[fail[u]][i];
      }
    }
  }
  void query(char* t) {
    int u = 0;
    for (int i = 1; t[i]; i++) {
      u = tr[u][t[i] - 'a'];
      ans[i] = exist[u];
    }
  }
} T1, T2;
char s[N];
char t[2000006];
int n;
int main() {
  scanf("%s", t + 1);
  int n;
  cin >> n;
  while (n--) {
    scanf("%s", s + 1);
    T1.insert(s);
    int l = strlen(s + 1);
    for (int i = 1; i <= l / 2; i++) swap(s[i], s[l - i + 1]);
    T2.insert(s);
  }
  T1.build();
  T2.build();
  T1.query(t);
  int l = strlen(t + 1);
  for (int i = 1; i <= l / 2; i++) swap(t[i], t[l - i + 1]);
  T2.query(t);
  long long res = 0;
  for (int i = 1; i <= l; i++) {
    res += 1ll * T1.ans[i] * T2.ans[l - i];
  }
  cout << res << ' ';
}

```