#include <bits/stdc++.h>
const int MAXN = 200010;
int qu[MAXN];
struct ACAM {
  int go[MAXN][26], fail[MAXN], flag[MAXN], cnt;
  ACAM() {
    memset(go, 0, sizeof go);
    memset(fail, 0, sizeof fail);
    memset(flag, 0, sizeof flag);
    cnt = 0;
  }
  void insert(char* s) {
    int now = 0;
    for (int i = 0; s[i]; ++i) {
      int u = s[i] - 'a';
      if (!go[now][u]) go[now][u] = ++cnt;
      now = go[now][u];
    }
    flag[now] += 1;
  }
  void build_up() {
    int hd = 0, tl = 0;
    for (int i = 0; i < 26; ++i)
      if (go[0][i]) qu[++tl] = go[0][i];
    while (hd < tl) {
      int now = qu[++hd];
      flag[now] += flag[fail[now]];
      for (int i = 0; i < 26; ++i)
        if (go[now][i])
          fail[go[now][i]] = go[fail[now]][i], qu[++tl] = go[now][i];
        else
          go[now][i] = go[fail[now]][i];
    }
    return;
  }
  inline int run(int& now, char a) {
    now = go[now][a - 'a'];
    return flag[now];
  }
} fwd, bck;
char t[MAXN], s[MAXN];
int n;
int fwd_mh[MAXN], bck_mh[MAXN];
int main() {
  scanf("%s%d", t, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s);
    fwd.insert(s);
    std::reverse(s, s + strlen(s));
    bck.insert(s);
  }
  fwd.build_up(), bck.build_up();
  int i;
  int now = 0;
  for (i = 0; t[i]; ++i) fwd_mh[i] = fwd.run(now, t[i]);
  now = 0;
  for (--i; ~i; --i) bck_mh[i] = bck.run(now, t[i]);
  long long int ans = 0;
  for (int i = 1; t[i]; ++i) ans += 1ll * fwd_mh[i - 1] * bck_mh[i];
  printf("%lld", ans);
  return 0;
}

### Prompt

Your challenge is to write 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>
const int MAXN = 200010;
int qu[MAXN];
struct ACAM {
  int go[MAXN][26], fail[MAXN], flag[MAXN], cnt;
  ACAM() {
    memset(go, 0, sizeof go);
    memset(fail, 0, sizeof fail);
    memset(flag, 0, sizeof flag);
    cnt = 0;
  }
  void insert(char* s) {
    int now = 0;
    for (int i = 0; s[i]; ++i) {
      int u = s[i] - 'a';
      if (!go[now][u]) go[now][u] = ++cnt;
      now = go[now][u];
    }
    flag[now] += 1;
  }
  void build_up() {
    int hd = 0, tl = 0;
    for (int i = 0; i < 26; ++i)
      if (go[0][i]) qu[++tl] = go[0][i];
    while (hd < tl) {
      int now = qu[++hd];
      flag[now] += flag[fail[now]];
      for (int i = 0; i < 26; ++i)
        if (go[now][i])
          fail[go[now][i]] = go[fail[now]][i], qu[++tl] = go[now][i];
        else
          go[now][i] = go[fail[now]][i];
    }
    return;
  }
  inline int run(int& now, char a) {
    now = go[now][a - 'a'];
    return flag[now];
  }
} fwd, bck;
char t[MAXN], s[MAXN];
int n;
int fwd_mh[MAXN], bck_mh[MAXN];
int main() {
  scanf("%s%d", t, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s);
    fwd.insert(s);
    std::reverse(s, s + strlen(s));
    bck.insert(s);
  }
  fwd.build_up(), bck.build_up();
  int i;
  int now = 0;
  for (i = 0; t[i]; ++i) fwd_mh[i] = fwd.run(now, t[i]);
  now = 0;
  for (--i; ~i; --i) bck_mh[i] = bck.run(now, t[i]);
  long long int ans = 0;
  for (int i = 1; t[i]; ++i) ans += 1ll * fwd_mh[i - 1] * bck_mh[i];
  printf("%lld", ans);
  return 0;
}

```