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 double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 100;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int maxm = 2e6 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
long long ans;
int cnt[maxn], tnc[maxn];
struct node {
  int trie[maxn][26];
  int num[maxn];
  int fail[maxn];
  int h[maxn];
  int vis[maxn];
  int last[maxn];
  int tot = 0;
  void init() { memset(vis, 0, sizeof(vis)); }
  void add(char *s) {
    int root = 0, len = strlen(s + 1);
    for (int i = 1; i <= len; i++) {
      int x = s[i] - 'a';
      if (!trie[root][x]) trie[root][x] = ++tot;
      root = trie[root][x];
      h[root] = i;
    }
    num[root]++;
  }
  void build() {
    queue<int> q;
    for (int i = 0; i < 26; i++) {
      if (trie[0][i]) {
        fail[trie[0][i]] = 0;
        q.push(trie[0][i]);
      }
    }
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      last[now] = num[fail[now]] ? fail[now] : last[fail[now]];
      for (int i = 0; i < 26; i++) {
        if (trie[now][i]) {
          fail[trie[now][i]] = trie[fail[now]][i];
          q.push(trie[now][i]);
        } else
          trie[now][i] = trie[fail[now]][i];
      }
    }
  }
  void acGao(char *s) {
    int root = 0;
    int len = strlen(s + 1);
    for (int i = 1; i <= len; i++) {
      int x = s[i] - 'a';
      root = trie[root][x];
      for (int j = root; j; j = last[j]) {
        cnt[i] += num[j];
        tnc[i - h[j] + 1] += num[j];
      }
    }
  }
} ac;
char S[maxn];
char a[maxn];
int n;
int main() {
  scanf("%s", S + 1);
  int m = strlen(S + 1);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", a + 1);
    int len = strlen(a + 1);
    ac.add(a);
  }
  ac.h[0] = 0;
  ac.build();
  ac.acGao(S);
  for (int i = 1; i <= m; i++) {
    ans += 1ll * cnt[i] * tnc[i + 1];
  }
  printf("%lld\n", ans);
  return 0;
}