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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define watch(x) cout<<(#x)<<"="<<(x)<<endl
#define mset(d,val) memset(d,val,sizeof(d))
#define setp(x) cout<<fixed<<setprecision(x)
#define forn(i,a,b) for(int i=a;i<b;i++)
#define fore(i,a,b) for(int i=a;i<=b;i++)
#define pb push_back
#define F first
#define S second
#define PI 3.14159265358979323846264338327
#define INF 2e14
#define MOD 998244353
#define pqueue priority_queue
#define fbo find_by_order
#define ook order_of_key
typedef long long ll;
typedef pair<ll,ll> ii;
typedef vector<ll> vi;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;

#define MAXN 100005

string s,t;
set<int> sa[26]; //locations of letters in s

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin>>s>>t;
	forn(i,0,s.length()){
		sa[s[i]-'a'].insert(i);
	}
	
	forn(i,0,t.length()){
		if(sa[t[i]-'a'].empty()){
			cout<<-1<<'\n';
			return 0;
		}
	}
	
	int p=-1,mul=0;
	
	/*forn(i,0,26){
		cout<<char('a'+i)<<':';
		for(set<int>::iterator it=sa[i].begin();it!=sa[i].end();it++) cout<<*it<<" ";
		cout<<'\n';
	}*/
	
	forn(i,0,t.length()){
		if(p>=0) p%=t.length();	//cout<<"t["<<i<<"]:";
		set<int>::iterator ptr=sa[t[i]-'a'].upper_bound(p);
		if(ptr==sa[t[i]-'a'].end()){ //cout<<"overflow...";
			mul++;
			ptr=sa[t[i]-'a'].lower_bound(0);
		}
		
		p=*ptr; //cout<<p<<'\n';
	}
	
	cout<<mul*s.length()+p+1<<'\n';
	
	return 0;
}