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:
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF1202E extends PrintWriter {
	CF1202E() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1202E o = new CF1202E(); o.main(); o.flush();
	}

	static final int A = 26;
	int[][] delta;
	int[] kk;
	void aho_corasick(byte[][] bb, int m) {
		int n = 1;
		for (int h = 0; h < m; h++)
			n += bb[h].length;
		kk = new int[n + 1];
		int[][] tt = new int[n + 1][A];
		n = 1;
		for (int h = 0; h < m; h++) {
			int s = 1, l = bb[h].length;
			for (int k = 0; k < l; k++) {
				int a = bb[h][k] - 'a';
				if (tt[s][a] == 0)
					tt[s][a] = ++n;
				s = tt[s][a];
			}
			kk[s]++;
		}
		delta = new int[n + 1][A];
		int[] qq = new int[n]; int head = 0, cnt = 0;
		int[] ff = new int[n + 1];
		ff[1] = 0; qq[head + cnt++] = 1;
		while (cnt > 0) {
			int s = qq[head++]; cnt--;
			int e = ff[s];
			kk[s] += kk[e];
			for (int a = 0; a < A; 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, byte[] aa, int n, byte[][] bb, int m) {
		aho_corasick(bb, m);
		int s = 1;
		for (int i = 0; i < n; i++) {
			int a = aa[i] - 'a';
			s = delta[s][a];
			pp[i] = kk[s];
		}
	}
	void reverse(byte[] cc, int n) {
		for (int i = 0, j = n - 1; i < j; i++, j--) {
			byte tmp = cc[i]; cc[i] = cc[j]; cc[j] = tmp;
		}
	}
	void main() {
		byte[] aa = sc.next().getBytes();
		int n = aa.length;
		int m = sc.nextInt();
		byte[][] bb = new byte[m][];
		for (int h = 0; h < m; h++)
			bb[h] = sc.next().getBytes();
		int[] pp = new int[n];
		solve(pp, aa, n, bb, m);
		reverse(aa, n);
		for (int h = 0; h < m; h++)
			reverse(bb[h], bb[h].length);
		int[] qq = new int[n];
		solve(qq, aa, n, bb, m);
		long ans = 0;
		for (int i = 0; i + 1 < n; i++)
			ans += (long) pp[i] * qq[n - 1 - (i + 1)];
		println(ans);
	}
}