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;
struct state {
  int nxt[26];
  int link, cnt;
  state() : link(0), cnt(0) { memset(nxt, 0, sizeof nxt); }
};
vector<state> tree;
int& nxt(int u, char c) { return tree[u].nxt[c - 'a']; }
int& link(int u) { return tree[u].link; }
int& cnt(int u) { return tree[u].cnt; }
void insert(const string& s) {
  int pt = 0;
  for (char c : s) {
    if (!nxt(pt, c)) {
      nxt(pt, c) = tree.size();
      tree.emplace_back();
    }
    pt = nxt(pt, c);
  }
  cnt(pt)++;
}
void automata() {
  queue<int> q;
  q.push(0);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    cnt(u) += cnt(link(u));
    for (char c = 'a'; c <= 'z'; c++) {
      int v = nxt(u, c);
      if (v) {
        link(v) = u ? nxt(link(u), c) : 0;
        q.push(v);
        nxt(u, c) = v;
      } else
        nxt(u, c) = nxt(link(u), c);
    }
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  string t;
  cin >> t;
  int m = t.size();
  vector<int> pre(m), suf(m);
  int k;
  cin >> k;
  vector<string> x(k);
  tree.emplace_back();
  for (string& s : x) {
    cin >> s;
    insert(s);
  }
  automata();
  int pt = 0;
  for (int i = 0; i < m; i++) {
    pt = nxt(pt, t[i]);
    pre[i] += cnt(pt);
  }
  tree.clear();
  tree.emplace_back();
  for (string& s : x) {
    reverse(s.begin(), s.end());
    insert(s);
  }
  automata();
  pt = 0;
  reverse(t.begin(), t.end());
  for (int i = 0; i < m; i++) {
    pt = nxt(pt, t[i]);
    suf[i] += cnt(pt);
  }
  reverse(suf.begin(), suf.end());
  long long res = 0;
  for (int i = 1; i < m; i++) res += 1ll * pre[i - 1] * suf[i];
  cout << res << '\n';
}