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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <queue>

#define rep(i,n) for(int i=0;i<n;i++)
#define dup(x,y) (((x)+(y)-1)/(y)) // dup * y >= x なる最小のdup.
using namespace std;
typedef long long ll;
using Graph = vector<vector<int>>;

int main() {
	string s, t;
	cin >> s >> t;
	// check: tに使われている文字が，sに含まれているか
	map<char, vector<ll>> s_dic;
	ll ssize = s.size();

	rep(i, ssize) {
		s_dic[s[i]].push_back(i);
	}
	rep(i, t.size()) {
		if(s_dic[t[i]].empty()) {
			cout << -1 << endl;
			return 0;
		} 
	}

	ll cnt = 0;
	//vector<ll> tmp;
	ll place = 0;
	rep(i, t.size()) {
		//cout << "search: " << t[i] << endl;
		//ll place = cnt%s.size();
		//tmp = s_dic[t[i]];
		if(place>s_dic[t[i]][s_dic[t[i]].size()-1]) {
			cnt += ssize - place;
			place = 0;
		}
		
		for(ll id : s_dic[t[i]]) {
			if(place<=id) {
				//cout << "id " << id << endl; 
				cnt += (id - place + 1);
				place = id+1;
				break;
			} 
		}
		
		//cout << "check " << cnt << endl;
	}

	cout << cnt << endl;
	return 0;
}