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;
char s[N], t[N];
int dp[2][N], q[N], n, m, len, i, j;
long long ans;
struct AC {
  int a[N][26], nxt[N], d[N], sz = 1;
  void ins(char s[N]) {
    int now = 1;
    for (int i = 0; s[i]; ++i) {
      if (!a[now][s[i] - 'a']) a[now][s[i] - 'a'] = ++sz;
      now = a[now][s[i] - 'a'];
    }
    d[now]++;
  }
  void acmach() {
    for (int i = 0; i < 26; ++i) a[0][i] = 1;
    int t = 0, w = 1;
    q[0] = 1;
    while (t != w) {
      int now = q[t++];
      for (int i = 0; i < 26; ++i)
        if (a[now][i]) {
          nxt[a[now][i]] = a[nxt[now]][i];
          d[a[now][i]] += d[nxt[a[now][i]]];
          q[w++] = a[now][i];
        } else
          a[now][i] = a[nxt[now]][i];
    }
  }
  void go(int dp[N]) {
    for (int now = 1, i = 0; i < n; ++i)
      now = a[now][s[i] - 'a'], dp[i] += d[now];
  }
} c1, c2;
int main() {
  for (scanf("%s%d", s, &m), n = strlen(s); m--;)
    scanf("%s", t), len = strlen(t), c1.ins(t), reverse(t, t + len), c2.ins(t);
  c1.acmach();
  c2.acmach();
  c1.go(dp[0]);
  reverse(s, s + n);
  c2.go(dp[1]);
  for (i = 1; i < n; ++i) ans += 1LL * dp[0][i - 1] * dp[1][n - i - 1];
  printf("%lld\n", ans);
}