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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 7;

char s[N], t[N];
int ne[N][26];
int fir[26];

int main() {
	scanf("%s%s", s + 1, t + 1);
	int len1 = strlen(s + 1), len2 = strlen(t + 1);
	for (int i = len1; i >= 0; i--) {
		for (int j = 0; j < 26; j++)
			ne[i][j] = fir[j];
		if (i) fir[s[i] - 'a'] = i;
	}
	int id = 0, ans = 1;
	for (int i = 1; i <= len2; i++) {
		if (ne[id][t[i] - 'a']) {
			id = ne[id][t[i] - 'a'];
		} else {
			if (!fir[t[i] - 'a']) {
				puts("-1");
				return 0;
			}
			id = fir[t[i] - 'a'];
			ans++;
		}
	}
	printf("%lld\n", 1LL * (ans - 1) * len1 + id);
	return 0;
}