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;
int g[200005][26], f[200005], w[200005], nc;
int gn() {
  int p = nc++;
  f[p] = w[p] = 0;
  memset(g[p], 0, sizeof(g[p]));
  return p;
}
void clr() {
  nc = 0;
  gn();
}
void ins(const string& s) {
  int p = 0;
  for (char ch : s) {
    int o = ch - 'a';
    if (!g[p][o]) g[p][o] = gn();
    p = g[p][o];
  }
  w[p]++;
}
void build() {
  queue<int> q;
  for (int o = 0; o != 26; ++o)
    if (g[0][o]) q.push(g[0][o]);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    w[u] += w[f[u]];
    for (int o = 0; o != 26; ++o) {
      int& v = g[u][o];
      if (!v)
        v = g[f[u]][o];
      else
        f[v] = g[f[u]][o], q.push(v);
    }
  }
}
char t[200005];
int ans1[200005], ans2[200005];
void match(int* ans) {
  int p = 0;
  for (int i = 0; t[i]; ++i) {
    int o = t[i] - 'a';
    p = g[p][o];
    ans[i] += w[p];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> t;
  int m = strlen(t);
  int n;
  cin >> n;
  vector<string> sv(n);
  for (int i = 0; i != n; ++i) cin >> sv[i];
  clr();
  for (int i = 0; i != n; ++i) ins(sv[i]);
  build();
  match(ans1);
  clr();
  reverse(t, t + m);
  for (int i = 0; i != n; ++i) {
    reverse(sv[i].begin(), sv[i].end());
    ins(sv[i]);
  }
  build();
  match(ans2);
  long long ans = 0;
  for (int i = 0; i != m - 1; ++i) ans += 1ll * ans1[i] * ans2[m - i - 2];
  cout << ans << endl;
  return 0;
}