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)

///A drunk man will find his way home, but a drunk bird may get lost forever...
#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define int long long
typedef long long ll;
typedef double dd;
typedef pair <int, int> pii;
#define pb push_back
#define pf push_front
#define ppb pop_back()
#define ppf pop_front()
#define f first
#define s second
#define sz(x) (int)x.size()
#define smin(x , y) x = min(x , y)
#define smax(x , y) x = max(x , y)
#define mp make_pair
#define all(x) x.begin() , x.end()
//const int mx = 1e5 + 5;
set < int > v[30];
set < pii > ms;
int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
	string s, t;
	//	ms.clear();
		cin >> s >> t;
	//	for(int i = 0; i < 30; i++)
	//		v[i].clear();
		for(int i = 0; i < sz(s); i++)
			v[s[i] - 'a'].insert(i);
		int l = 0, ans = 0, ind = -1;
		bool b = 0;
		for(int i = 0; i < sz(t); i++){
			int x = t[i] - 'a';
	//		cout << i << " " << x << " " << ind << endl;
			if(ms.find({x , ind}) != ms.end()){
		//		cout << "vcbhxdjvhkjdfvhkjhgfk " << ind << endl;
				if(v[x].upper_bound(ind) != v[x].end()){
		//			cout << "bro " << endl;
					ind = *v[x].upper_bound(ind);
			//		cout << "ind1 " << ind << endl;
					l++;
					ms.insert({x , ind});
				}
				else{
					if(!l){
	//					cout << "i " << i << endl;
						b = 1;
						break;
					}
		//			cout << "l1 " << l << endl;
					ms.clear();
					l = 0;
					ans++;
					ind = 0;
					i--;
				}
			}
			else{
		//		cout << "blah" << endl;
				if(v[x].lower_bound(ind) != v[x].end()){
					ind = *v[x].lower_bound(ind);
				//	cout << "ind2 " << ind << endl;
					l++;
					ms.insert({x , ind});
				}
				else{
					if(!l){
			//			cout << "i2 " << i << endl;
						b = 1;
						break;
					}
				//	cout << "l2 " << l << endl;
					ms.clear();
					l = 0;
					ans++;
					ind = 0;
					i--;
				}
			}
		}
		if(b)
			cout << -1 << endl;
		else
			cout << ans * sz(s) + (l > 0 ? ind + 1 : 0) << endl;
	return 0;
}