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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll =long long;
using P =pair<int,int>;

int main(){
    string s,t;
    int l;
    cin >> s >> t;
    l=s.size();
    vector <vector <int>> al(26);

    rep(i,26){
        al[i].push_back(-1);
    }
    rep(i,l){
        al[s[i]-97].push_back(i);
    }


    ll ans=0;
    int k=0;
    int l2;
    int rt,lt,m;
    int l3;
    l3=t.size();

    rep(i,l3){
        l2=al[t[i]-97].size();
        //cout << l2 << endl;
        if(l2==1){
            ans=-1;
            break;
        }
        if(k>al[t[i]-97][l2-1]){
            ans+=l;
            k=0;
        }
        rt=l2-1;
        lt=0;
        while(rt-lt>1){
            m=(rt+lt)/2;
            if(al[t[i]-97][m]<=k){
                lt=m;
            }
            else{
                rt=m;
            }
        }
        k=al[t[i]-97][rt];
        ++k;
        //cout << k << endl;
        if(i==(l3-1)){
            ans+=k;
        }


    }

    
    cout << ans << endl;
    

    return 0;
}