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 JAVA):

Solution:
import java.util.ArrayDeque;
import java.util.Queue;
import java.util.Scanner;

public class cf_1202E {
    static class ACA {
        static class Node {
            Node next[], fail;
            int end;
            Node() {
                next = new Node[26];
            }
        }
        Node root;
        ACA() {
            root = new Node();
        }
        void insert(String str, int len) {
            Node cur = root;
            for(int i = 0; i < len; i++) {
                int ch = str.charAt(i) - 'a';
                if(cur.next[ch] == null) {
                    cur.next[ch] = new Node();
                }
                cur = cur.next[ch];
            }
            cur.end++;
        }
        void build() {
            Queue<Node> que = new ArrayDeque<>();
            root.fail = root;
            for(int i = 0; i < 26; i++) {
                if(root.next[i] == null) {
                    root.next[i] = root;
                }
                else {
                    root.next[i].fail = root;
                    que.add(root.next[i]);
                }
            }
            while(!que.isEmpty()) {
                Node u = que.poll();
                u.end += u.fail.end;
                for(int i = 0; i < 26; i++) {
                    if(u.next[i] == null) {
                        u.next[i] = u.fail.next[i];
                    }
                    else {
                        u.next[i].fail = u.fail.next[i];
                        que.add(u.next[i]);
                    }
                }
            }
        }
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        String t = in.next();
        int n = in.nextInt();
        String v[] = new String[n];
        int cnt[] = new int[t.length()];
        ACA aca = new ACA();

        for(int i = 0; i < n; i++) {
            v[i] = in.next();
            aca.insert(v[i], v[i].length());
        }
        aca.build();
        ACA.Node cur = aca.root;
        for(int i = 0; i < t.length(); i++) {
            cur = cur.next[t.charAt(i) - 'a'];
            cnt[i] = cur.end;
        }

        long ans = 0;
        aca = new ACA();
        for(int i = 0; i < n; i++) {
            aca.insert(new StringBuilder(v[i]).reverse().toString(), v[i].length());
        }
        aca.build();
        cur = aca.root;
        for(int i = t.length() - 1; i >= 1; i--) {
            cur = cur.next[t.charAt((i)) - 'a'];
            ans += (long)cnt[i - 1] * cur.end;
        }

        System.out.println(ans);
    }
}