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 <iostream>
#include <string>
using namespace std;

string s, t;
long long m;
long long p[100000];

int main()
{	
	cin >> s >> t;

	long long w = s.size();

	s = s + s;

	for (int i = 0; i < t.size(); i++) {
		if (s.find_first_of(t[i]) == std::string::npos) {
			m = -1;
			break;
		}
		else {
			if (i == 0) {
				p[0] = s.find_first_of(t[0]);
				m = p[0] + 1;
			}
			else {
				long long u = s.find_first_of(t[i], p[i - 1] + 1);

				p[i] = u % w;
				m += u - p[i - 1];
			}
		}
	}

	cout << m << endl;

	return 0;
}