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>
void reverse(char *cc) {
  int n = strlen(cc), i, j;
  for (i = 0, j = n - 1; i < j; i++, j--) {
    char tmp;
    tmp = cc[i], cc[i] = cc[j], cc[j] = tmp;
  }
}
int kk[200001 + 1], delta[200001 + 1][26];
void aho_corasick(char *bb[], int m) {
  static int tt[200001 + 1][26], ff[200001 + 1], qq[200001 + 1];
  int n, h, s, head, cnt;
  n = 1;
  for (h = 0; h < m; h++) n += strlen(bb[h]);
  memset(kk, 0, (n + 1) * sizeof *kk);
  for (s = 1; s <= n; s++) memset(tt[s], 0, sizeof tt[s]);
  n = 1;
  for (h = 0; h < m; h++) {
    int k, l = strlen(bb[h]);
    s = 1;
    for (k = 0; k < l; k++) {
      int a = bb[h][k] - 'a';
      if (!tt[s][a]) tt[s][a] = ++n;
      s = tt[s][a];
    }
    kk[s]++;
  }
  head = cnt = 0;
  ff[1] = 0, qq[head + cnt++] = 1;
  while (cnt) {
    int e, a;
    s = qq[cnt--, head++];
    e = ff[s];
    kk[s] += kk[e];
    for (a = 0; a < 26; a++) {
      int t = tt[s][a];
      int f = e == 0 ? 1 : delta[e][a];
      if (t == 0)
        delta[s][a] = f;
      else {
        delta[s][a] = t;
        ff[t] = f;
        qq[head + cnt++] = t;
      }
    }
  }
}
void solve(int *pp, char *aa, int n, char *bb[], int m) {
  int s, i;
  aho_corasick(bb, m);
  s = 1;
  for (i = 0; i < n; i++) {
    int a = aa[i] - 'a';
    s = delta[s][a];
    pp[i] = kk[s];
  }
}
int main() {
  static char aa[200000 + 1], *bb[200000];
  static int pp[200000], qq[200000];
  int n, m, h, i;
  long long ans;
  scanf("%s", aa), n = strlen(aa);
  scanf("%d", &m);
  for (h = 0; h < m; h++) {
    static char s[200000 + 1];
    scanf("%s", s);
    bb[h] = strdup(s);
  }
  solve(pp, aa, n, bb, m);
  reverse(aa);
  for (h = 0; h < m; h++) reverse(bb[h]);
  solve(qq, aa, n, bb, m);
  ans = 0;
  for (i = 0; i + 1 < n; i++) ans += (long long)pp[i] * qq[n - 1 - (i + 1)];
  printf("%lld\n", ans);
  return 0;
}