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: Wrong Answer)

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;

int main(){
    string s;
    string t;
    cin >> s;
    cin >> t;
    int N=s.size();
    s+=s; //sの長さを二倍にしておく
    vector<int> v[50];
    for(int i=0;i<s.size();i++){
        v[s[i]-'a'].push_back(i); //sにおける各アルファベットのindexを追加
    }
    long long ans=0;
    int ind=0;
    int cnt=0;
    for(int i=0;i<t.size();i++){
        if(v[t[i]-'a'].size()==0){
            cout << -1 << endl;
            return 0;
        }
        int tmp=*lower_bound(v[t[i]-'a'].begin(),v[t[i]-'a'].end(),ind);
        if(tmp>=N){
            cnt++;
            tmp-=N;
        }
        ind=tmp;
    }
    ans=ind+1+cnt*N;
    cout << ans << endl;

}