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>
using namespace std;

int main()
{
  string s, t;
  cin >> s >> t;

  vector<bool> table('z'-'a'+1, false);
  vector<vector<long long>> m('z'-'a'+1);
  for (long long i = 0; i < (long long)s.size(); i++) {
    char c = s.at(i);
    table.at(c-'a') = true;
    m.at(c-'a').push_back(i);
  }
  bool can_be_solved = true;
  for (char c : t) {
    if (!table.at(c-'a')) {
      can_be_solved = false;
    }
  }
  if (!can_be_solved) {
    cout << "-1\n";
    return 0;
  }

  long long cnt = 0;
  long long idx = -1;
  for (char c : t) {
    auto it = upper_bound(m.at(c-'a').begin(), m.at(c-'a').end(), idx);
    if (it == m.at(c-'a').end()) {
      idx = -1;
      cnt++;
      it = upper_bound(m.at(c-'a').begin(), m.at(c-'a').end(), idx);
    }
    idx = *it;
  }
  cout << s.size() * cnt + idx + 1 << '\n';

  return 0;
}