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>
#define ll long long
using namespace std;

int main(){
    ll i, j;
    string s, t;
    cin >> s >> t;
    vector<ll> c_idx[26];
    for(i = 0; i < s.length(); i++){
        c_idx[s[i] - 'a'].push_back(i);
    }
    ll next[26][s.length()];
    for(i = 0; i < 26; i++){
        ll k = 0;
        if(c_idx[i].size() == 0)
            continue;
        for(j = 0; j < c_idx[i].size(); j++){
            ll t = c_idx[i][j];
            while(k < t){
                next[i][k] = t;
                k++;
            }
        }
        while(k < s.length()){
            next[i][k] = c_idx[i][0];
            k++;
        }
    }
    ll a = 0;
    j = s.length() - 1;
    for(i = 0; i < t.length(); i++){
        if(c_idx[t[i] - 'a'].size() == 0){
            a = -1;
            break;
        }
        ll n = next[t[i] - 'a'][j];
        if(n <= j){
            a += n + s.length() - j;
        }
        else{
            a += n - j;
        }
        j = n;
    }
    cout << a << endl;
}