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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
string tt, s[200005];
int n, c[200005];
long long e[200005], st[200005];
const int K = 26;
map<string, int> m;
struct Vertex {
  int next[K];
  bool leaf = false;
  int p = -1;
  char pch;
  int link = -1;
  int exit_link = -1;
  int go[K];
  vector<int> id;
  Vertex(int p = -1, char ch = '$') : p(p), pch(ch) {
    fill(begin(next), end(next), -1);
    fill(begin(go), end(go), -1);
  }
};
vector<Vertex> t(1);
void add_string(int i, int keep = 1) {
  int v = 0;
  for (char ch : s[i]) {
    int c = ch - 'a';
    if (t[v].next[c] == -1) {
      t[v].next[c] = t.size();
      t.emplace_back(v, ch);
    }
    v = t[v].next[c];
  }
  t[v].leaf = true;
  if (keep) t[v].id.push_back(i);
}
int go(int v, char ch);
int get_link(int v) {
  if (t[v].link == -1) {
    if (v == 0 || t[v].p == 0)
      t[v].link = 0;
    else
      t[v].link = go(get_link(t[v].p), t[v].pch);
  }
  return t[v].link;
}
int get_exit_link(int v) {
  if (t[v].exit_link == -1) {
    if (v == 0 || t[v].p == 0)
      t[v].exit_link = 0;
    else {
      int temp = get_link(v);
      if (t[temp].leaf)
        t[v].exit_link = temp;
      else
        t[v].exit_link = get_exit_link(temp);
    }
  }
  return t[v].exit_link;
}
int go(int v, char ch) {
  int c = ch - 'a';
  if (t[v].go[c] == -1) {
    if (t[v].next[c] != -1)
      t[v].go[c] = t[v].next[c];
    else
      t[v].go[c] = v == 0 ? 0 : go(get_link(v), ch);
  }
  return t[v].go[c];
}
int main() {
  cin >> tt >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    if (m.find(s[i]) != m.end())
      add_string(i, 0);
    else {
      m[s[i]] = i;
      add_string(i);
    }
    c[m[s[i]]]++;
  }
  int v = 0;
  for (int i = 0; i < tt.size(); i++) {
    v = go(v, tt[i]);
    int temp = v;
    if (!t[temp].leaf) temp = get_exit_link(temp);
    while (temp) {
      for (auto j : t[temp].id) {
        st[i - s[j].size() + 1] += c[j];
        e[i] += c[j];
      }
      temp = get_exit_link(temp);
    }
  }
  long long ans = 0;
  for (int i = 0; i < tt.size(); i++) ans += e[i] * st[i + 1];
  cout << ans << "\n";
}