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 <iostream>
#include <algorithm>
#include <string>
using namespace std;
typedef long long ll;


int main(void){
  int n,m,i,j,l[27][100005]={};
  string s,t;
  
  cin >> s >> t;
  n=s.size();
  m=t.size();
  
  for (i=0;i<n;i++) l[s[i]-'a'][i]++;
  for (i=0;i<26;i++) for (j=1;j<=n;j++){
    l[i][j]+=l[i][j-1];
  }
  
  for (i=0;i<m;i++){
    if (l[t[i]-'a'][n]==0){
      cout << -1 << endl;
      return 0;
    }
  }
  
  int p=0,lk,rou=0;
  int *sp;
  for (i=0;i<m;i++){
    lk=t[i]-'a';
    sp=lower_bound(l[lk]+p,l[lk]+n,l[lk][p]+1);
    if (p>=n || sp==l[lk]+n){
      p=0;
      sp=lower_bound(l[lk],l[lk]+n,1);
      rou++;
    }
    p=sp-l[lk];
  }
  
  cout << n*rou+p+1 << endl;
  
  return 0;
}