#include <bits/stdc++.h>
using namespace std;
long long n;
struct trie {
  long long tot = 0, ch[400011][26], val[400011], fail[400011], ans[400011];
  void insert(string s) {
    long long x = 0;
    for (long long i = 0; i < s.size(); i++) {
      long long d = s[i] - 'a';
      if (!ch[x][d]) ch[x][d] = ++tot;
      x = ch[x][d];
    }
    val[x]++;
  }
  void build() {
    queue<long long> q;
    for (long long i = 0; i < 26; i++) {
      if (ch[0][i]) {
        q.push(ch[0][i]);
      }
    }
    long long sum = 0;
    while (!q.empty()) {
      long long x = q.front();
      q.pop();
      for (long long i = 0; i < 26; i++) {
        if (ch[x][i]) {
          fail[ch[x][i]] = ch[fail[x]][i];
          val[ch[x][i]] += val[fail[ch[x][i]]];
          q.push(ch[x][i]);
        } else
          ch[x][i] = ch[fail[x]][i];
      }
    }
  }
  void query(string s) {
    long long x = 0;
    for (long long i = 0; i < s.size(); i++) {
      x = ch[x][s[i] - 'a'];
      ans[i] = val[x];
    }
  }
} accm, raccm;
signed main() {
  string t;
  cin >> t;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    string s;
    cin >> s;
    long long len = s.size();
    accm.insert(s);
    reverse(s.begin(), s.end());
    raccm.insert(s);
  }
  accm.build();
  raccm.build();
  accm.query(t);
  long long len = t.size();
  reverse(t.begin(), t.end());
  raccm.query(t);
  reverse(raccm.ans, raccm.ans + len);
  long long ans = 0;
  for (long long i = 0; i < len; i++) {
    ans += accm.ans[i] * raccm.ans[i + 1];
  }
  cout << ans;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
long long n;
struct trie {
  long long tot = 0, ch[400011][26], val[400011], fail[400011], ans[400011];
  void insert(string s) {
    long long x = 0;
    for (long long i = 0; i < s.size(); i++) {
      long long d = s[i] - 'a';
      if (!ch[x][d]) ch[x][d] = ++tot;
      x = ch[x][d];
    }
    val[x]++;
  }
  void build() {
    queue<long long> q;
    for (long long i = 0; i < 26; i++) {
      if (ch[0][i]) {
        q.push(ch[0][i]);
      }
    }
    long long sum = 0;
    while (!q.empty()) {
      long long x = q.front();
      q.pop();
      for (long long i = 0; i < 26; i++) {
        if (ch[x][i]) {
          fail[ch[x][i]] = ch[fail[x]][i];
          val[ch[x][i]] += val[fail[ch[x][i]]];
          q.push(ch[x][i]);
        } else
          ch[x][i] = ch[fail[x]][i];
      }
    }
  }
  void query(string s) {
    long long x = 0;
    for (long long i = 0; i < s.size(); i++) {
      x = ch[x][s[i] - 'a'];
      ans[i] = val[x];
    }
  }
} accm, raccm;
signed main() {
  string t;
  cin >> t;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    string s;
    cin >> s;
    long long len = s.size();
    accm.insert(s);
    reverse(s.begin(), s.end());
    raccm.insert(s);
  }
  accm.build();
  raccm.build();
  accm.query(t);
  long long len = t.size();
  reverse(t.begin(), t.end());
  raccm.query(t);
  reverse(raccm.ans, raccm.ans + len);
  long long ans = 0;
  for (long long i = 0; i < len; i++) {
    ans += accm.ans[i] * raccm.ans[i + 1];
  }
  cout << ans;
}

```