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 N = 2e5 + 7;
using ll = long long;
string t;
int c[N], revc[N], nxt[N];
vector<string> ls;
struct Trie {
  int nxt[N][26], tot, rt, cnt[N], fail[N];
  void init() {
    tot = 0;
    rt = newnode();
  }
  int newnode() {
    ++tot;
    memset(nxt[tot], 0, sizeof(nxt[tot]));
    cnt[tot] = 0;
    return tot;
  }
  void insert(const string &s) {
    int now = rt;
    for (int i = 0; i < s.length(); ++i) {
      int ch = s[i] - 'a';
      if (nxt[now][ch] == 0) nxt[now][ch] = newnode();
      now = nxt[now][ch];
    }
    ++cnt[now];
  }
  void build() {
    queue<int> q;
    for (int i = 0; i < 26; ++i) {
      if (nxt[rt][i] == 0)
        nxt[rt][i] = rt;
      else {
        fail[nxt[rt][i]] = rt;
        q.push(nxt[rt][i]);
      }
    }
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      cnt[now] += cnt[fail[now]];
      for (int i = 0; i < 26; ++i) {
        if (nxt[now][i] == 0)
          nxt[now][i] = nxt[fail[now]][i];
        else {
          fail[nxt[now][i]] = nxt[fail[now]][i];
          q.push(nxt[now][i]);
        }
      }
    }
  }
  void solve(int *c) {
    int now = rt;
    for (int i = 0; i < t.length(); ++i) {
      int ch = t[i] - 'a';
      now = nxt[now][ch];
      c[i] += cnt[now];
    }
  }
} trie;
void cal(int *c) {
  trie.init();
  for (string &s : ls) trie.insert(s);
  trie.build();
  trie.solve(c);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    ls.push_back(s);
  }
  cal(c);
  reverse(t.begin(), t.end());
  for (string &s : ls) reverse(s.begin(), s.end());
  cal(revc);
  ll ans = 0;
  for (int i = 0; i + 1 < t.length(); ++i) {
    ans += 1LL * c[i] * revc[t.length() - i - 2];
  }
  cout << ans << endl;
}