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 mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
long long gcd(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
long long ksm(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = (r * a) % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return r;
}
const int maxn = 2e5 + 100;
int n, m, k;
struct ACM {
  queue<int> q;
  int cnt = 0, trie[maxn][26], fail[maxn], sum[maxn], num[maxn];
  void insert(string &s) {
    int now = 0;
    for (int i = 0; i < s.size(); i++) {
      int u = s[i] - 'a';
      if (!trie[now][u]) trie[now][u] = ++cnt;
      now = trie[now][u];
    }
    sum[now]++;
  }
  void build() {
    while (!q.empty()) q.pop();
    for (int i = 0; i < 26; i++)
      if (trie[0][i]) q.push(trie[0][i]);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int i = 0; i < 26; i++) {
        if (trie[u][i]) {
          fail[trie[u][i]] = trie[fail[u]][i];
          sum[trie[u][i]] += sum[fail[trie[u][i]]];
          q.push(trie[u][i]);
        } else
          trie[u][i] = trie[fail[u]][i];
      }
    }
  }
  void solve(string &s) {
    int now = 0;
    for (int i = 0; i < s.size(); i++) {
      int u = s[i] - 'a';
      now = trie[now][u];
      num[i] = sum[now];
    }
  }
} AC[2];
string t, s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> t >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    AC[0].insert(s);
    reverse(s.begin(), s.end());
    AC[1].insert(s);
  }
  AC[0].build();
  AC[1].build();
  AC[0].solve(t);
  reverse(t.begin(), t.end());
  AC[1].solve(t);
  long long ans = 0;
  for (int i = 0; i < t.size(); i++) {
    ans += (long long)AC[0].num[i] * AC[1].num[t.size() - 2 - i];
  }
  cout << ans << endl;
}