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;
vector<int>g[30];
int main(){
  string s, t;
  long long ans=0, c;
  cin>>s>>t;
  for(int i=0;i<s.length();i++){
    g[s[i]-'a'].push_back(i+1);
  }
  vector<int>::iterator it;
  for(int i=0;i<30;i++)g[i].push_back(1000000);
  ans+=g[t[0]-'a'][0];
  c=ans;
  for(int i=1;i<t.length();i++){
    int f=t[i]-'a';
    if(g[f][0]==1000000){
      cout<<-1<<endl;
      exit(0);
    }
    it=upper_bound(g[f].begin(), g[f].end(), c);
    if(*it==1000000){
      ans+=g[f][0];
      ans+=(s.length()-c);
      c=g[f][0];
    }
    else {
      ans+=(*it-c);
      c=*it;
    }
  }
  cout<<ans<<endl;
}