#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int n;
char s[200005], ch[200005];
long long ans;
struct AC {
  int cnt, bj[200005], fail[200005], ch[200005][26], g[200005];
  void Insert(char *s, int len) {
    int now = 1;
    assert(strlen(s) == len);
    for (int i = 0; i < len; ++i) {
      int &o = ch[now][s[i] - 'a'];
      if (!o) o = ++cnt;
      now = o;
    }
    bj[now]++;
  }
  void Build() {
    queue<int> q;
    for (int i = 0; i <= 25; ++i) ch[0][i] = 1;
    q.push(1);
    while (!q.empty()) {
      int x = q.front(), f = fail[x];
      q.pop();
      bj[x] += bj[f];
      for (int i = 0; i <= 25; ++i) {
        int &y = ch[x][i];
        if (!y)
          y = ch[f][i];
        else
          fail[y] = ch[f][i], q.push(y);
      }
    }
  }
  void query(char *s) {
    int len = strlen(s), now = 1;
    for (int i = 0; i < len; ++i) {
      now = ch[now][s[i] - 'a'];
      g[i] = bj[now];
    }
  }
} A, B;
int main() {
  scanf("%s", s);
  n = read();
  A.cnt = B.cnt = 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%s", ch);
    int len = strlen(ch);
    A.Insert(ch, len);
    reverse(ch, ch + len);
    B.Insert(ch, len);
  }
  int len = strlen(s);
  A.Build();
  A.query(s);
  reverse(s, s + len);
  B.Build();
  B.query(s);
  for (int i = 0; i < len - 1; ++i) ans += 1ll * A.g[i] * B.g[len - i - 2];
  cout << ans << '\n';
}

### Prompt

Create a solution in CPP for 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;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int n;
char s[200005], ch[200005];
long long ans;
struct AC {
  int cnt, bj[200005], fail[200005], ch[200005][26], g[200005];
  void Insert(char *s, int len) {
    int now = 1;
    assert(strlen(s) == len);
    for (int i = 0; i < len; ++i) {
      int &o = ch[now][s[i] - 'a'];
      if (!o) o = ++cnt;
      now = o;
    }
    bj[now]++;
  }
  void Build() {
    queue<int> q;
    for (int i = 0; i <= 25; ++i) ch[0][i] = 1;
    q.push(1);
    while (!q.empty()) {
      int x = q.front(), f = fail[x];
      q.pop();
      bj[x] += bj[f];
      for (int i = 0; i <= 25; ++i) {
        int &y = ch[x][i];
        if (!y)
          y = ch[f][i];
        else
          fail[y] = ch[f][i], q.push(y);
      }
    }
  }
  void query(char *s) {
    int len = strlen(s), now = 1;
    for (int i = 0; i < len; ++i) {
      now = ch[now][s[i] - 'a'];
      g[i] = bj[now];
    }
  }
} A, B;
int main() {
  scanf("%s", s);
  n = read();
  A.cnt = B.cnt = 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%s", ch);
    int len = strlen(ch);
    A.Insert(ch, len);
    reverse(ch, ch + len);
    B.Insert(ch, len);
  }
  int len = strlen(s);
  A.Build();
  A.query(s);
  reverse(s, s + len);
  B.Build();
  B.query(s);
  for (int i = 0; i < len - 1; ++i) ans += 1ll * A.g[i] * B.g[len - i - 2];
  cout << ans << '\n';
}

```