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: Wrong Answer)

import java.util.Scanner;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		String s=sc.next();
		String t=sc.next();
		int[][] kk=new int[26][s.length()];
		for(int i=0; i<26; i++){
			int k=s.length();
			for(int j=s.length()-1; j>=0; j--){
				if(s.charAt(j)-'a'==i){
					if(j==0){
						k=j;
					}else{
						kk[i][j]=k;
						k=j;
						continue;
					}
				}
				kk[i][j]=k;
			}
		}
		int c=0;
		int count=0;
		for(int i=0; i<t.length(); i++){
			if(c==0 && kk[t.charAt(i)-'a'][0]==s.length()){
				System.out.println(-1);
				return;
			}
			c=kk[t.charAt(i)-'a'][c];
			if(c==s.length()){
				count++;
				c=0;
				i--;
			}
		}
		System.out.println(count*s.length()+c+1);
	}
}