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 scan = new Scanner(System.in);
		String s = scan.nextLine();
		String t = scan.nextLine();
		int len = t.length();

		// 最大値を超えるような条件は与えられない
//		double max = Math.pow(10, 100);

		String tmpS = "";
		char[] tChars = t.toCharArray();
		int lastIndex = -1;
		int tmpIndex = -1;
		double loopCnt = 0;
		double ans = -1;
		int cnt = 0;

		for (char c : tChars) {
			// 続きから後ろを確認
			tmpIndex = s.indexOf(c, lastIndex + 1);
			if (tmpIndex != -1) {
				lastIndex = tmpIndex;
			}
			else {
				// 見つからなければ、次週に
				loopCnt++;
//				if (loopCnt >= max) {
//					break;
//				}
				tmpS = s.substring(0, lastIndex + 1);
				lastIndex = tmpS.indexOf(c);
				// 前から探しても見つからなければ、その文字は無いため終了
				if (lastIndex == -1) {
					break;
				}
			}
			cnt++;
			if (cnt == len) {
				ans = (s.length() * loopCnt) + lastIndex + 1;
			}
		}
		System.out.println((int)ans);
	}
}