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 ACAM {
  int tree[200005][26], fail[200005], tot, head[200005], pnt[200005],
      nxt[200005];
  int endpos[200005], E;
  void insert(string s) {
    int u = 0, len = s.length();
    for (int i = 0; i < len; i++) {
      if (!tree[u][s[i] - 'a']) tree[u][s[i] - 'a'] = ++tot;
      u = tree[u][s[i] - 'a'];
    }
    endpos[u]++;
  }
  void build() {
    queue<int> q;
    for (int i = 0; i < 26; i++)
      if (tree[0][i]) q.push(tree[0][i]);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int i = 0; i < 26; i++) {
        if (!tree[u][i])
          tree[u][i] = tree[fail[u]][i];
        else {
          fail[tree[u][i]] = tree[fail[u]][i];
          q.push(tree[u][i]);
        }
      }
    }
  }
  void add_edge(int u, int v) {
    pnt[E] = v;
    nxt[E] = head[u];
    head[u] = E++;
  }
  void init() {
    memset(head, -1, sizeof(head));
    for (int i = 1; i <= tot; i++) add_edge(fail[i], i);
  }
  void dfs(int u, int pre, int w) {
    endpos[u] += w;
    w = endpos[u];
    for (int i = head[u]; i != -1; i = nxt[i]) {
      int v = pnt[i];
      if (v == pre) continue;
      dfs(v, u, w);
    }
  }
} AC1, AC2;
string t;
int n, a[200005];
int main() {
  cin >> t >> n;
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    AC1.insert(s);
    reverse(s.begin(), s.end());
    AC2.insert(s);
  }
  AC1.build();
  AC2.build();
  AC1.init();
  AC2.init();
  AC1.dfs(0, 0, 0);
  AC2.dfs(0, 0, 0);
  int u = 0;
  for (int i = 0; i < t.length(); i++) {
    u = AC1.tree[u][t[i] - 'a'];
    a[i] = AC1.endpos[u];
  }
  long long ans = 0;
  u = 0;
  reverse(t.begin(), t.end());
  for (int i = 0; i < (int)t.length() - 1; i++) {
    u = AC2.tree[u][t[i] - 'a'];
    ans += 1ll * a[t.length() - i - 2] * AC2.endpos[u];
  }
  cout << ans;
}