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 <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    string s, t;

    cin >> s;
    cin >> t;

    vector<vector<int>> apparent(26);

    for (int i = 0; i < s.size(); i++) {
        apparent[s[i] - 'a'].emplace_back(i);
    }

    int last_idx = 0;
    int ans = 0;

    for (int i = 0; i < t.size(); i++) {
        int poss = upper_bound(apparent[t[i] - 'a'].begin(), apparent[t[i] - 'a'].end(), last_idx) - apparent[t[i] - 'a'].begin();
        if (poss == apparent[t[i] - 'a'].size()) {
            if (poss == 0) {
                cout << "-1"
                     << "\n";
                return 0;
            } else {
                ans += s.size() - last_idx + apparent[t[i] - 'a'][poss - 1];
                last_idx = apparent[t[i] - 'a'][poss - 1];
                //cout << ans << "\n";
            }
        } else {
            ans += apparent[t[i] - 'a'][poss] - last_idx;
            last_idx = apparent[t[i] - 'a'][poss];
            //cout << ans << "\n";
        }
    }

    cout << ans + 1 << "\n";
    return 0;
}