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;
const int B = 300;
using ll = long long;
string t;
int c[N], revc[N], nxt[N];
vector<string> ls, ss;
struct Trie {
  int nxt[N][26], tot, rt, cnt[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];
  }
  int query(const string &t, int start) {
    int res = 0;
    int now = rt;
    for (int i = start; i < t.length(); ++i) {
      int ch = t[i] - 'a';
      now = nxt[now][ch];
      if (now == 0) break;
      res += cnt[now];
    }
    return res;
  }
} trie;
void kmp_pre(const string &s) {
  int i, j;
  j = nxt[0] = -1;
  i = 0;
  while (i < s.length()) {
    while (-1 != j && s[i] != s[j]) j = nxt[j];
    nxt[++i] = ++j;
  }
}
void kmp(const string &s, int *c) {
  int i = 0, j = 0, ans = 0;
  kmp_pre(s);
  while (i < t.length()) {
    while (-1 != j && t[i] != s[j]) j = nxt[j];
    ++i;
    ++j;
    if (j >= s.length()) {
      ++c[i - s.length()];
      j = nxt[j];
    }
  }
}
void cal(int *c) {
  trie.init();
  for (string &s : ss) {
    trie.insert(s);
  }
  for (int i = 0; i < t.length(); ++i) {
    c[i] = trie.query(t, i);
  }
  for (string &s : ls) {
    kmp(s, 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;
    if (s.length() < B)
      ss.push_back(s);
    else
      ls.push_back(s);
  }
  cal(c);
  reverse(t.begin(), t.end());
  for (string &s : ls) reverse(s.begin(), s.end());
  for (string &s : ss) reverse(s.begin(), s.end());
  cal(revc);
  ll ans = 0;
  for (int i = 1; i < t.length(); ++i) {
    ans += 1LL * c[i] * revc[t.length() - i];
  }
  cout << ans << endl;
}