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;
const int K = 26;
struct Vertex {
  int next[K];
  long long int leaf = 0;
  int p = -1;
  char pch;
  int link = -1;
  int go[K];
  long long int count = -1;
  int leaflink = 0;
  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(string const& s) {
  int v = 0;
  for (char ch : s) {
    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++;
}
int go(int v, char ch);
int get_link(int v);
long long int cnt(int v) {
  if (t[v].count == -1) {
    if (v == 0)
      t[v].count = 0;
    else
      t[v].count = cnt(get_link(v)) + t[v].leaf;
  }
  return t[v].count;
}
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 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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string ts;
  cin >> ts;
  int n;
  cin >> n;
  vector<string> v;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    v.push_back(s);
    reverse(s.begin(), s.end());
    add_string(s);
  }
  int m = ts.size();
  vector<long long int> a = vector<long long int>(m, 0);
  int cs = 0;
  for (int i = m - 1; i > -1; --i) {
    get_link(cs);
    cs = go(cs, ts[i]);
    get_link(cs);
    a[i] = cnt(cs);
  }
  t = vector<Vertex>(1);
  for (int i = 0; i < n; ++i) {
    add_string(v[i]);
  }
  cs = 0;
  long long int sol = 0;
  for (int i = 0; i < m - 1; ++i) {
    get_link(cs);
    cs = go(cs, ts[i]);
    get_link(cs);
    sol += cnt(cs) * a[i + 1];
  }
  cout << sol << endl;
}