#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
template <typename T>
void read(T& x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  x *= f;
}
int n;
char s[N], t[N];
int f[N][2];
struct AC {
  int ac[N][26], fail[N], ends[N];
  int tot;
  void insert(char* s) {
    int len = strlen(s + 1);
    int u = 0;
    for (int i = 1; i <= len; i++) {
      int j = s[i] - 'a';
      if (!ac[u][j]) ac[u][j] = ++tot;
      u = ac[u][j];
    }
    ends[u]++;
  }
  void build() {
    queue<int> q;
    for (int i = 0; i <= 25; i++)
      if (ac[0][i]) q.push(ac[0][i]);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      ends[u] += ends[fail[u]];
      for (int i = 0; i <= 25; i++)
        if (ac[u][i])
          fail[ac[u][i]] = ac[fail[u]][i], q.push(ac[u][i]);
        else
          ac[u][i] = ac[fail[u]][i];
    }
  }
} t1, t2;
int main() {
  scanf("%s", t + 1);
  read(n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    t1.insert(s);
    int len = strlen(s + 1);
    reverse(s + 1, s + len + 1);
    t2.insert(s);
  }
  t1.build();
  t2.build();
  int u = 0, len = strlen(t + 1);
  for (int i = 1; i <= len; i++) {
    u = t1.ac[u][t[i] - 'a'];
    f[i][0] = t1.ends[u];
  }
  reverse(t + 1, t + len + 1);
  u = 0;
  for (int i = 1; i <= len; i++) {
    u = t2.ac[u][t[i] - 'a'];
    f[i][1] = t2.ends[u];
  }
  long long ans = 0;
  for (int i = 1; i <= len; i++) ans += 1ll * f[i][0] * f[len - i][1];
  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;
template <typename T>
void read(T& x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  x *= f;
}
int n;
char s[N], t[N];
int f[N][2];
struct AC {
  int ac[N][26], fail[N], ends[N];
  int tot;
  void insert(char* s) {
    int len = strlen(s + 1);
    int u = 0;
    for (int i = 1; i <= len; i++) {
      int j = s[i] - 'a';
      if (!ac[u][j]) ac[u][j] = ++tot;
      u = ac[u][j];
    }
    ends[u]++;
  }
  void build() {
    queue<int> q;
    for (int i = 0; i <= 25; i++)
      if (ac[0][i]) q.push(ac[0][i]);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      ends[u] += ends[fail[u]];
      for (int i = 0; i <= 25; i++)
        if (ac[u][i])
          fail[ac[u][i]] = ac[fail[u]][i], q.push(ac[u][i]);
        else
          ac[u][i] = ac[fail[u]][i];
    }
  }
} t1, t2;
int main() {
  scanf("%s", t + 1);
  read(n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    t1.insert(s);
    int len = strlen(s + 1);
    reverse(s + 1, s + len + 1);
    t2.insert(s);
  }
  t1.build();
  t2.build();
  int u = 0, len = strlen(t + 1);
  for (int i = 1; i <= len; i++) {
    u = t1.ac[u][t[i] - 'a'];
    f[i][0] = t1.ends[u];
  }
  reverse(t + 1, t + len + 1);
  u = 0;
  for (int i = 1; i <= len; i++) {
    u = t2.ac[u][t[i] - 'a'];
    f[i][1] = t2.ends[u];
  }
  long long ans = 0;
  for (int i = 1; i <= len; i++) ans += 1ll * f[i][0] * f[len - i][1];
  printf("%lld\n", ans);
  return 0;
}

```