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)

//  Created by conan1024hao in 2019.
//  Copyright © 2019 conan1024hao. All rights reserved.
//  専用ライブラリです、自由にコピーして構いません。
//  感谢看我的代码！Wechat:conan1024hao QQ:810396815
#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define INF 10e16
#define MOD 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
using Int=int_fast64_t;
typedef long long int ll;
typedef pair<ll,ll>Pll;
typedef pair<int,int>Pin;
long long GCD(long long a, long long b) { return b ? GCD(b, a%b) : a; }
long long LCM(long long a, long long b)  {return a/GCD(a,b)*b;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
ll cmp(pair<ll,string>a,pair<ll,string> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se<b.se;
}

//---------------------------------------------------------------------------

//---------------------------------------------------------------------------
int main(){//問題をちゃんと見ろ!!!!!!!!!!!!!!!!!      llか??????????
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------
    //ex:conteSt contEst coNTEst coNtest ContE
    string s,t;cin>>s>>t;
    vector<int>alp[26];map<char,int>mm;
    for(int i=0;i<s.length();i++){
            alp[s[i]-'a'].pb(i);
    }
    unsigned long long cnt1=0;
    if(alp[t[0]-'a'].empty()){
        cout<<-1<<endl;
        return 0;
    }
    int now=-1;
    for(int i=0;i<t.length();){
        if(alp[t[i]-'a'].empty()){
            cout<<-1<<endl;
            return 0;
        }
        bool abc=0;
        //cout<<now<<endl;
        for(int j=0;j<alp[t[i]-'a'].size();j++){
            
            //cout<<now<<" "<<alp[t[i]-'a'][j]<<endl;
            if(now<alp[t[i]-'a'][j]){
                now=alp[t[i]-'a'][j];
                abc=1;
                i++;
                break;
            }
        }
        if(abc==0){
            cnt1++;
            now=-1;
        }
        //if(now==alp[t[i]-'a'][0])
        /*if(alp[t[i]-'a'][]<now){
            cnt1++;
            m.clear();
            if(i<t.length()-1){
            if(alp[t[i+1]-'a'].empty()){
                cout<<-1<<endl;
                return 0;
            }
            now=alp[t[i]-'a'][0];
            //m[t[i]-'a']++;
               // i++;
            }
        }
        else{
            now=alp[t[i]-'a'][m[t[i]-'a']];
            m[t[i]-'a']++;
        }
         */
        //cout<<cnt1<<endl;
    }
    cout<<(unsigned long long)(cnt1*s.length()+now+1)<<endl;
    //-------------------------------
    return 0;
}
//---------------------------------------------------------------------------