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)

//In the name of God!
#include <bits/stdc++.h>
const int maxn = 2e5 + 12;
using namespace std;
vector<int> v[26];
string s, t;
int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    cin >> s >> t;
    for(int i = 0; i < s.size(); i++)
        v[s[i] - 'a'].push_back(i);

    long long ans = 0, i = 0;
    bool b;
    for(int ind = 0;ind < t.size(); ind++){
        if(v[t[ind] - 'a'].empty()){
            cout << -1;
            return 0;
        }
        b = 0;
        int j = lower_bound(v[t[ind] - 'a'].begin(), v[t[ind] - 'a'].end(), i) - v[t[ind] - 'a'].begin();
        j = v[t[ind] - 'a'][j];
        if(j >= i){
            ans += j + 1 - i;
            b = 1;
            i = j + 1;
        }
        if(!b){
            ans += s.size() - i + v[t[ind] - 'a'][0] + 1;
            i = v[t[ind] - 'a'][0] + 1;
        }
    }
    cout << ans;
}