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: Time Limit Exceeded)

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

using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  string s, t;
  cin >> s >> t;
  vi t_i(t.length() + 1, 0);
  string ss = s + s;
  for (int i = 0; i < t.length(); i++) {
    for (int j = t_i[i]; j < ss.length(); j++) {
      if (t[i] == ss[j]) {
        t_i[i + 1] = (j >= s.length()) ? j - s.length() + 1 : j + 1;
        break;
      }
    }
    if (t_i[i + 1] == 0) {
      cout << -1 << endl;
      return 0;
    }
  }
  // for (int i = 0; i < t_i.size(); i++) {
  //   printf("%d ", t_i[i]);
  // }
  // printf("\n");

  ll ans = 0;
  int l = 1;
  int r = 1;
  for (int i = 1; i < t_i.size(); i++) {
    if (i == t_i.size() - 1) {
      // printf("last\n");
      ans += t_i[i];
    } else {
      if (t_i[i] >= t_i[i + 1]) {
        ans += s.length();
        // printf("ans=%d\n", ans);
      }
    }
  }
  cout << ans << endl;
  return 0;
}