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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = 1e18;

signed main(){
	string s, t; cin >> s >> t;
	vector<vector<ll> > data(26);
	vector<int> cs(26, 0), ct(26, 0);
	ll n = s.size(), m = t.size();
	for(ll i = 0; i < n; i++) data[s[i] - 'a'].push_back(i), cs[s[i] - 'a']++;
	for(int i = 0; i < m; i++) ct[t[i] - 'a']++;
	for(ll i = 0; i < 26; i++){
		if(cs[i] == 0 && ct[i] != 0){
			cout << -1 << endl;
			return 0;
		}
	}
	for(int i = 0; i < 26; i++) data[i].push_back(INF);
	
	ll ans = 0, now = 0;
	for(ll i = 0; i < m; i++){
		ll num = t[i] - 'a';
		now = *upper_bound(data[num].begin(), data[num].end(), now);
		if(now == INF) now = data[num][0], ans++;
	}
	ans = ans * n + now;
	if(now != 0) ans++;
	cout << ans << endl;
}