#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
char t[N], s[N];
struct Aho {
  int trie[N][26];
  int cnt[N];
  int fail[N];
  int tot = 0;
  int dp[N];
  void init(int u) {
    for (int i = 0; i < 26; i++) trie[u][i] = 0;
    fail[u] = cnt[u] = 0;
  }
  void add(char s[]) {
    int rt = 0, len = strlen(s);
    for (int i = 0; i < len; i++) {
      int c = s[i] - 'a';
      if (!trie[rt][c]) init(trie[rt][c] = ++tot);
      rt = trie[rt][c];
    }
    cnt[rt]++;
  }
  void getFail() {
    queue<int> q;
    for (int i = 0; i < 26; i++)
      if (trie[0][i]) {
        fail[trie[0][i]] = 0;
        q.push(trie[0][i]);
      }
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      cnt[cur] += cnt[fail[cur]];
      for (int i = 0; i < 26; i++) {
        if (trie[cur][i]) {
          fail[trie[cur][i]] = trie[fail[cur]][i];
          q.push(trie[cur][i]);
        } else
          trie[cur][i] = trie[fail[cur]][i];
      }
    }
  }
  void query(char s[]) {
    int rt = 0, len = strlen(s);
    for (int i = 0; i <= len - 1; i++) {
      rt = trie[rt][s[i] - 'a'];
      dp[i] = cnt[rt];
    }
  }
} ac1, ac2;
int main() {
  while (~scanf("%s%d", t, &n)) {
    ac1.init(ac1.tot = 0);
    ac2.init(ac2.tot = 0);
    for (int i = 1; i <= n; i++) {
      scanf("%s", s);
      ac1.add(s);
      reverse(s, s + strlen(s));
      ac2.add(s);
    }
    ac1.getFail();
    ac1.query(t);
    reverse(t, t + strlen(t));
    ac2.getFail();
    ac2.query(t);
    long long ans = 0LL;
    int len = strlen(t);
    for (int i = 0; i <= len - 2; i++)
      ans += 1LL * ac1.dp[i] * ac2.dp[len - i - 2];
    printf("%lld\n", ans);
  }
  return 0;
}

### 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 + 5;
int n;
char t[N], s[N];
struct Aho {
  int trie[N][26];
  int cnt[N];
  int fail[N];
  int tot = 0;
  int dp[N];
  void init(int u) {
    for (int i = 0; i < 26; i++) trie[u][i] = 0;
    fail[u] = cnt[u] = 0;
  }
  void add(char s[]) {
    int rt = 0, len = strlen(s);
    for (int i = 0; i < len; i++) {
      int c = s[i] - 'a';
      if (!trie[rt][c]) init(trie[rt][c] = ++tot);
      rt = trie[rt][c];
    }
    cnt[rt]++;
  }
  void getFail() {
    queue<int> q;
    for (int i = 0; i < 26; i++)
      if (trie[0][i]) {
        fail[trie[0][i]] = 0;
        q.push(trie[0][i]);
      }
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      cnt[cur] += cnt[fail[cur]];
      for (int i = 0; i < 26; i++) {
        if (trie[cur][i]) {
          fail[trie[cur][i]] = trie[fail[cur]][i];
          q.push(trie[cur][i]);
        } else
          trie[cur][i] = trie[fail[cur]][i];
      }
    }
  }
  void query(char s[]) {
    int rt = 0, len = strlen(s);
    for (int i = 0; i <= len - 1; i++) {
      rt = trie[rt][s[i] - 'a'];
      dp[i] = cnt[rt];
    }
  }
} ac1, ac2;
int main() {
  while (~scanf("%s%d", t, &n)) {
    ac1.init(ac1.tot = 0);
    ac2.init(ac2.tot = 0);
    for (int i = 1; i <= n; i++) {
      scanf("%s", s);
      ac1.add(s);
      reverse(s, s + strlen(s));
      ac2.add(s);
    }
    ac1.getFail();
    ac1.query(t);
    reverse(t, t + strlen(t));
    ac2.getFail();
    ac2.query(t);
    long long ans = 0LL;
    int len = strlen(t);
    for (int i = 0; i <= len - 2; i++)
      ans += 1LL * ac1.dp[i] * ac2.dp[len - i - 2];
    printf("%lld\n", ans);
  }
  return 0;
}

```