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 <algorithm>
#include <iostream>
#include <map>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;

int main(void) {
  ios::sync_with_stdio(false);
  string S, T;
  cin >> S >> T;

  map<char, vector<int>> mp;
  for (uint i = 0; i < S.size(); i++) {
    mp[S[i]].push_back(i + 1);
  }

  uint j = 0;
  int64_t Ans = 0;
  for (uint i = 0; i < T.size(); i++) {
    if (mp[T[i]].size() == 0) {
      cout << "-1" << endl;
      return 0;
    }
    auto itr = upper_bound(mp[T[i]].begin(), mp[T[i]].end(), j);
    if (itr != mp[T[i]].end()) {
      Ans += (*itr - j);
    } else {
      itr = mp[T[i]].begin();
      Ans += S.size() - j + *itr;
    }
    j = *itr;
  }
  cout << Ans << endl;
  return 0;
}