Problem:
Score : 500 points

Problem Statement

Given are two strings s and t consisting of lowercase English letters. Determine if there exists an integer i satisfying the following condition, and find the minimum such i if it exists.

Let s' be the concatenation of 10^{100} copies of s. t is a subsequence of the string {s'}_1{s'}_2\ldots{s'}_i (the first i characters in s').

Notes

A subsequence of a string a is a string obtained by deleting zero or more characters from a and concatenating the remaining characters without changing the relative order. For example, the subsequences of contest include net, c, and contest.

Constraints

1 \leq |s| \leq 10^5

1 \leq |t| \leq 10^5

s and t consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If there exists an integer i satisfying the following condition, print the minimum such i; otherwise, print -1.

Sample Input 1

contest
son

Sample Output 1

10

t = son is a subsequence of the string contestcon (the first 10 characters in s' = contestcontestcontest...), so i = 10 satisfies the condition.

On the other hand, t is not a subsequence of the string contestco (the first 9 characters in s'), so i = 9 does not satisfy the condition.

Similarly, any integer less than 9 does not satisfy the condition, either. Thus, the minimum integer i satisfying the condition is 10.

Sample Input 2

contest
programming

Sample Output 2

-1

t = programming is not a substring of s' = contestcontestcontest.... Thus, there is no integer i satisfying the condition.

Sample Input 3

contest
sentence

Sample Output 3

33

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

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

char s[114514];
char t[114514];

int mofu[26];
int* nyan[26];

int main(void) {
	int i;
	int slen;
	int zyun = 0, seek = 0;
	if (scanf("%114513s", s) != 1) return 1;
	if (scanf("%114513s", t) != 1) return 1;
	for (i = 0; s[i] != '\0'; i++) {
		int idx = s[i] - 'a';
		if (0 <= idx && idx < 26) {
			nyan[idx] = realloc(nyan[idx], sizeof(*nyan[idx]) * (mofu[idx] + 1));
			nyan[idx][mofu[idx]++] = i;
		}
	}
	slen = i;
	for (i = 0; t[i] != '\0'; i++) {
		int idx = t[i] - 'a';
		if (0 <= idx && idx < 26) {
			if (mofu[idx] <= 0) {
				puts("-1");
				return 0;
			}
			if (nyan[idx][mofu[idx] - 1] < seek) {
				zyun++;
				seek = nyan[idx][0] + 1;
			} else if (seek <= nyan[idx][0]) {
				seek = nyan[idx][0] + 1;
			} else {
				int less = 0, ge = mofu[idx] - 1;
				while (less + 1 < ge) {
					int mid = less + (ge - less) / 2;
					if (seek <= nyan[idx][mid]) ge = mid; else less = mid;
				}
				seek = ge + 1;
			}
			if (seek >= slen) {
				zyun++;
				seek = 0;
			}
		} else {
			puts("-1");
			return 0;
		}
	}
	printf("%" PRId64 "\n", (int64_t)zyun * slen + seek);
	return 0;
}