#include <bits/stdc++.h>
using namespace std;
int ch[200010][26], tot, n, al[200010], ed[200010];
int ins(const char* S, bool op = 0) {
  int x = 0;
  if (op)
    for (int i = strlen(S) - 1; ~i; i--)
      x = (ch[x][S[i] - 'a'] ? ch[x][S[i] - 'a'] : ch[x][S[i] - 'a'] = ++tot);
  else
    for (int i = 0; S[i]; i++)
      x = (ch[x][S[i] - 'a'] ? ch[x][S[i] - 'a'] : ch[x][S[i] - 'a'] = ++tot);
  return x;
}
void byd() {
  queue<int> Q;
  for (int i = 0; i < 26; i++)
    if (ch[0][i]) Q.push(ch[0][i]);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = 0; i < 26; i++)
      if (!ch[x][i])
        ch[x][i] = ch[al[x]][i];
      else
        al[ch[x][i]] = ch[al[x]][i], Q.push(ch[x][i]),
        ed[ch[x][i]] += ed[ch[al[x]][i]];
  }
}
void Mapping(const char* S, int* f, int op = 0) {
  int x = 0;
  if (op)
    for (int i = strlen(S) - 1; ~i; i--) x = ch[x][S[i] - 'a'], f[i] = ed[x];
  else
    for (int i = 0; S[i]; i++) x = ch[x][S[i] - 'a'], f[i] = ed[x];
}
char S[200010], T[200010];
long long ans;
int f[200010], g[200010];
string ss[200010];
signed main() {
  int x, y;
  scanf("%s%d", T, &n);
  for (int i = 1; i <= n; i++) scanf("%s", S), ss[i] = S, ++ed[ins(S)];
  byd(), Mapping(T, f);
  memset(ch, 0, sizeof(ch)), memset(ed, 0, sizeof(ed)),
      memset(al, 0, sizeof(al)), tot = 0;
  for (int i = 1; i <= n; i++) ++ed[ins(ss[i].c_str(), 1)];
  byd(), Mapping(T, g, 1);
  for (int i = 0, tt = strlen(T); i < tt; i++) ans += 1ll * f[i] * g[i + 1];
  cout << ans << '\n';
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 ch[200010][26], tot, n, al[200010], ed[200010];
int ins(const char* S, bool op = 0) {
  int x = 0;
  if (op)
    for (int i = strlen(S) - 1; ~i; i--)
      x = (ch[x][S[i] - 'a'] ? ch[x][S[i] - 'a'] : ch[x][S[i] - 'a'] = ++tot);
  else
    for (int i = 0; S[i]; i++)
      x = (ch[x][S[i] - 'a'] ? ch[x][S[i] - 'a'] : ch[x][S[i] - 'a'] = ++tot);
  return x;
}
void byd() {
  queue<int> Q;
  for (int i = 0; i < 26; i++)
    if (ch[0][i]) Q.push(ch[0][i]);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = 0; i < 26; i++)
      if (!ch[x][i])
        ch[x][i] = ch[al[x]][i];
      else
        al[ch[x][i]] = ch[al[x]][i], Q.push(ch[x][i]),
        ed[ch[x][i]] += ed[ch[al[x]][i]];
  }
}
void Mapping(const char* S, int* f, int op = 0) {
  int x = 0;
  if (op)
    for (int i = strlen(S) - 1; ~i; i--) x = ch[x][S[i] - 'a'], f[i] = ed[x];
  else
    for (int i = 0; S[i]; i++) x = ch[x][S[i] - 'a'], f[i] = ed[x];
}
char S[200010], T[200010];
long long ans;
int f[200010], g[200010];
string ss[200010];
signed main() {
  int x, y;
  scanf("%s%d", T, &n);
  for (int i = 1; i <= n; i++) scanf("%s", S), ss[i] = S, ++ed[ins(S)];
  byd(), Mapping(T, f);
  memset(ch, 0, sizeof(ch)), memset(ed, 0, sizeof(ed)),
      memset(al, 0, sizeof(al)), tot = 0;
  for (int i = 1; i <= n; i++) ++ed[ins(ss[i].c_str(), 1)];
  byd(), Mapping(T, g, 1);
  for (int i = 0, tt = strlen(T); i < tt; i++) ans += 1ll * f[i] * g[i + 1];
  cout << ans << '\n';
  return 0;
}

```