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: Runtime Error)

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

long long int rimac(const char *s1, const char *s2) {
	int i, j, k;
	const char *big, *small;
	(strlen(s1)-strlen(s2)>=0) ? (big=s1, small=s2) : (big=s2, small=s1);
	for (i=0; i<strlen(big)-strlen(small)+1; i++) {
		if (small[0] == big[i]) {
			for (j=i+1; j<strlen(big)-strlen(small)+2; j++) {
				if (small[1] == big[j]) {
					if (strcmp(&small[2], &big[j+1]) == 0) {
						return 1;
					}
				}
			}
		}
	}
	return 0;
}

int main(int argc, char **argv) {
	int i, j;
	int n;
	long long int count = 0;
	char **s;
	scanf("%d", &n);
	if ((s = (char **) calloc(n, sizeof(char *))) == NULL)
	{
		exit(1);
	}

	for (i=0; i<n; i++) {
		char buf[2000000];
		scanf("%s", buf);
		if ((s[i] = (char *)calloc(strlen(buf), sizeof(char)))
			== NULL) {
				exit(1);
		}
		strcpy(s[i], buf);	
	}

	for (i=0; i<n-1; i++) {
		for (j=i+1; j<n; j++) {
			count += rimac(s[i], s[j]);
		}
	}

	printf("%lld", count);
	return 0;
}