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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct node {
  int go[26], end;
  void init() {
    memset(go, -1, sizeof(go));
    end = 0;
  }
};
const int N = 2e5 + 42;
int n = 0, sq = 0, cnte[N] = {0}, cnts[N] = {0}, sz = 1;
node trie[N];
string t;
vector<string> s;
void add(string& s, int v, int i = 0) {
  if (i == s.size()) {
    trie[v].end++;
    return;
  }
  if (trie[v].go[s[i] - 'a'] == -1) {
    trie[v].go[s[i] - 'a'] = sz++;
    trie[sz - 1].init();
  }
  add(s, trie[v].go[s[i] - 'a'], i + 1);
}
int get(int v, int i) {
  cnte[i - 1] += trie[v].end;
  if (i != t.size() && trie[v].go[t[i] - 'a'] != -1)
    return trie[v].end + get(trie[v].go[t[i] - 'a'], i + 1);
  return trie[v].end;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  trie[0].init();
  cin >> t >> n;
  s.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    sq += s[i].size();
  }
  sq = sqrt(sq) + 1;
  for (auto& cur : s) {
    if (cur.size() <= sq)
      add(cur, 0);
    else {
      vector<int> b(cur.size() + 1);
      int i = 0;
      int j = b[0] = -1;
      while (i < cur.size()) {
        while (j != -1 && cur[i] != cur[j]) j = b[j];
        i++;
        j++;
        b[i] = j;
      }
      i = j = 0;
      while (i < t.size()) {
        while (j != -1 && t[i] != cur[j]) j = b[j];
        i++;
        j++;
        if (j == cur.size()) {
          cnte[i - 1]++;
          cnts[i - cur.size()]++;
          j = b[j];
        }
      }
    }
  }
  for (int i = 0; i < t.size(); i++) cnts[i] += get(0, i);
  long long ans = 0;
  for (int i = 0; i + 1 < t.size(); i++) ans += 1ll * cnte[i] * cnts[i + 1];
  cout << ans;
  return 0;
}