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>
using namespace std;
typedef long long ll;
#define INF 1LL<<62
#define inf 1000000007

vector<vector<ll>>a(26);
int main() {
	string s,t;
	cin>>s>>t;
	for(int i=0;i<s.size();i++){
		ll now=s[i]-'a';
		a[now].push_back(i+1);
	}
	ll ans=0,now=0;
	for(int i=0;i<t.size();i++){
		ll sear=t[i]-'a';
		bool ch=false;
		for(int j=0;j<a[sear].size();j++){
			ll fi =a[sear][j];
			if(fi>now){
				ans+=fi-now;
				now=fi;
				ch=true;
				break;
			}
		}
		if(ch!=true){
			i--;
			ans+=s.size()-now;
			now=0;
		}
	}
	cout << ans;
	return 0;
}