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>
#define rep(i,n) for(int i=0; i<(int)(n); i++)

using namespace std;
typedef long long LL;

int main(){
  string s, t;
  cin >> s >> t;
  map<char,vector<int>> m;
  rep(i,s.size()){
    char c=s[i];
    m[c].emplace_back(i);
  }
  LL num=-1, pos=0, ans=0;
  LL n=s.size();
  rep(i,t.size()){
    char c=t[i];
    if(!m[c].size()){
      cout << -1 << endl;
      return 0;
    }
    auto itr=upper_bound(m[c].begin(),m[c].end(),num);
    if(itr-m[c].begin()==m[c].size()){
      pos++;
      num=m[c][0];
      ans=m[c][0]+n*pos;
    }
    else{
      num=*itr;
      ans=num+n*pos;
    }
  }
  cout << ans+1 << endl;

  return 0;
}