#include <bits/stdc++.h>
using namespace std;
int n, tot[2], ch[2][210000][26], fa[2][210000], fail[2][210000];
long long Ans, sum[2][210000], ans[2][210000];
void addfail(int m, char *f, int len) {
  int now = 0;
  for (int i = 0; i < len; i++) {
    if (!ch[m][now][f[i] - 'a']) ++tot[m], ch[m][now][f[i] - 'a'] = tot[m];
    now = ch[m][now][f[i] - 'a'];
  }
  sum[m][now]++;
}
void bfs(int m) {
  queue<int> q;
  sum[m][0] = 0;
  for (int i = 0; i < 26; i++)
    if (ch[m][0][i]) q.push(ch[m][0][i]);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < 26; i++) {
      int v = ch[m][u][i];
      if (!v)
        ch[m][u][i] = ch[m][fail[m][u]][i];
      else
        fail[m][v] = ch[m][fail[m][u]][i], q.push(v);
    }
    sum[m][u] += sum[m][fail[m][u]];
  }
}
void getans(int m, char *f, int len) {
  int now = 0;
  for (int i = 0; i < len; i++)
    now = ch[m][now][f[i] - 'a'], ans[m][i] = sum[m][now];
}
char a[210000], s[210000];
int main() {
  scanf("%s", a);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    int len = strlen(s);
    addfail(0, s, len), reverse(s, s + len), addfail(1, s, len);
  }
  int len = strlen(a);
  bfs(0), bfs(1), getans(0, a, len), reverse(a, a + len), getans(1, a, len);
  reverse(ans[1], ans[1] + len);
  for (int i = 0; i < len - 1; i++) Ans += ans[0][i] * ans[1][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;
int n, tot[2], ch[2][210000][26], fa[2][210000], fail[2][210000];
long long Ans, sum[2][210000], ans[2][210000];
void addfail(int m, char *f, int len) {
  int now = 0;
  for (int i = 0; i < len; i++) {
    if (!ch[m][now][f[i] - 'a']) ++tot[m], ch[m][now][f[i] - 'a'] = tot[m];
    now = ch[m][now][f[i] - 'a'];
  }
  sum[m][now]++;
}
void bfs(int m) {
  queue<int> q;
  sum[m][0] = 0;
  for (int i = 0; i < 26; i++)
    if (ch[m][0][i]) q.push(ch[m][0][i]);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < 26; i++) {
      int v = ch[m][u][i];
      if (!v)
        ch[m][u][i] = ch[m][fail[m][u]][i];
      else
        fail[m][v] = ch[m][fail[m][u]][i], q.push(v);
    }
    sum[m][u] += sum[m][fail[m][u]];
  }
}
void getans(int m, char *f, int len) {
  int now = 0;
  for (int i = 0; i < len; i++)
    now = ch[m][now][f[i] - 'a'], ans[m][i] = sum[m][now];
}
char a[210000], s[210000];
int main() {
  scanf("%s", a);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    int len = strlen(s);
    addfail(0, s, len), reverse(s, s + len), addfail(1, s, len);
  }
  int len = strlen(a);
  bfs(0), bfs(1), getans(0, a, len), reverse(a, a + len), getans(1, a, len);
  reverse(ans[1], ans[1] + len);
  for (int i = 0; i < len - 1; i++) Ans += ans[0][i] * ans[1][i + 1];
  printf("%lld\n", Ans);
  return 0;
}

```