Problem:
Score : 700 points

Problem Statement

Limak can repeatedly remove one of the first two characters of a string,
for example abcxyx \rightarrow acxyx \rightarrow cxyx \rightarrow cyx.

You are given N different strings S_1, S_2, \ldots, S_N.
Among N \cdot (N-1) / 2 pairs (S_i, S_j),
in how many pairs could Limak obtain one string from the other?

Constraints

2 \leq N \leq 200\,000

S_i consists of lowercase English letters a-z.

S_i \neq S_j

1 \leq |S_i|

|S_1| + |S_2| + \ldots + |S_N| \leq 10^6

Input

Input is given from Standard Input in the following format.

N
S_1
S_2
\vdots
S_N

Output

Print the number of unordered pairs (S_i, S_j)
where i \neq j and Limak can obtain one string from the other.

Sample Input 1

3
abcxyx
cyx
abc

Sample Output 1

1

The only good pair is (abcxyx, cyx).

Sample Input 2

6
b
a
abc
c
d
ab

Sample Output 2

5

There are five good pairs: (b, abc), (a, abc), (abc, c), (b, ab), (a, ab).

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

#define HASH 200003

typedef struct List {
	struct List *next;
	char *s;
	int l;
} list;

long long div_mod(long long x, long long y, long long z)
{
	if (x % y == 0) return x / y;
	else return (div_mod((1 + x / y) * y - x, (z % y), y) * z + x) / y;
}

int hash_string(char* s)
{
	int i, x = 0;
	for (i = 0; s[i] != 0; i++) x = ((x << 5) + s[i] - 'a') % HASH;
	return x;
}

int lex_smaller(char* s1, char* s2)
{
	int i;
	for (i = 0; s1[i] != 0 && s2[i] != 0; i++) {
		if (s1[i] < s2[i]) return 1;
		else if (s1[i] > s2[i]) return -1;
	}
	if (s1[i] == s2[i]) return 0;
	else if (s1[i] == 0) return 1;
	else return -1;
}

int main()
{
	int i, j, l[200001], N;
	char *S[200001], T[1000001];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) {
		scanf("%s", T);
		for (l[i] = 0; T[l[i]] != 0; l[i]++);
		S[i] = (char*)malloc(sizeof(char) * (l[i] + 1));
		for (j = 0; j <= l[i]; j++) S[i][j] = T[j];
	}
	
	int h[200001];
	list *p, *hash[HASH], d[200001];
	for (i = 1; i <= N; i++) {
		h[i] = hash_string(S[i]);
		d[i].s = S[i];
		d[i].l = l[i];
		d[i].next = hash[h[i]];
		hash[h[i]] = &(d[i]);
	}
	
	int k, dp[26];
	long long ans = 0, tmp, pow[32];
	const long long inv = div_mod(1, 32, HASH);
	for (k = 1, pow[0] = 32; k < 32; k++) pow[k] = pow[k-1] * pow[k-1] % HASH;
	for (i = 1; i <= N; i++) {
		if (l[i] <= 1) continue;
		for (j = 0; j <= l[i]; j++) T[j] = S[i][j];
		for (j = l[i] - 1, k = 0, tmp = 1; j > 0; j >>= 1, k++) if ((j & 1) == 1) tmp = tmp * pow[k] % HASH;
		h[i] = (h[i] - tmp * (T[0] - 'a') % HASH + HASH) % HASH;
		tmp = tmp * inv % HASH;
		h[i] = (h[i] - tmp * (T[1] - 'a') % HASH + HASH) % HASH;
		for (j = 0; j < 26; j++) dp[j] = 0;
		dp[S[i][0] - 'a'] = 1;
		for (j = 1; j < l[i]; j++) {
			dp[S[i][j] - 'a'] = 1;
			for (k = 0; k < 26; k++) {
				if (dp[k] == 0) continue;
				T[j] = 'a' + k;
				h[0] = (tmp * k + h[i]) % HASH;
				for (p = hash[h[0]]; p != NULL; p = p->next) if (p->l == l[i] - j) break;
				if (p != NULL) ans++;
			}
			tmp = tmp * inv % HASH;
			h[i] = (h[i] - tmp * (S[i][j+1] - 'a') % HASH + HASH) % HASH;
		}
	}
	
	printf("%lld\n", ans);
	fflush(stdout);
	return 0;
}