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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main{

	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
		char[] s = scn.nextLine().toCharArray();
		char[] t = scn.nextLine().toCharArray();
		int N = s.length;
		boolean[] se = new boolean[26], te = new boolean[26];
		for(int i = 0;i < s.length;i++) {
			int a = 'a';
			se[s[i]-a] = true;
		}
		for(int i = 0;i < t.length;i++) {
			int a = 'a';
			te[t[i]-a] = true;
		}
		boolean can = true;
		for(int i = 0;i < 26;i++) {
			can = se[i] || !te[i];
			if(!can)break;
		}

		long[][] next = new long[N][26];

		if(can) {

			for(int i = 0;i < 26;i++) {
				int now = 'a'+i;
				int count = 0;
				boolean first = false;
				for(int j = 0;j < N;j++) {
					if(s[j] == now) {
						if(j == 0) {
							first = true;
							count++;
							continue;
						}else {
							for(int k = j-count;k < j;k++) {
								next[k][i] = j-k;
							}
							count=1;
						}
					}else {
						count++;
					}
				}
				if(first) {
					for(int k = N-count;k < N;k++) {
						next[k][i] = N-k;
					}
				}else {
					for(int k = N-count;k < N;k++) {
						next[k][i] = N-k+next[0][i];
					}
				}
			}
			long ans = 0;
			int i = 0;
			if(t[0]==s[0])i=1;
			for(;i < t.length;i++) {
				int a = 'a';
				ans += next[(int)((ans)%N)][t[i]-a];
			}

			System.out.println(ans+1);

		}else {
			System.out.println(-1);
		}
	}

}