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 <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
using namespace std ;
string s , t ;
long long sum = 0 ; 
vector<int>c[30] ;
int main () {
    cin >> s >> t ;
    int n = s.size() ;
    int m = t.size() ;
    for(int i = 0 ; i < s.size() ; i ++) {
        c[s[i]-'a'].push_back(i+1) ;
    }
    int x = t[0] - 'a' , j ;
    if(!c[x].size()) {
        puts("-1") ;
        return 0 ;
    }
    int k = c[x][0] ;
    sum += k ;
    for(int i = 1 ; i < m ; i ++) {
        x = t[i] - 'a' ;
        if(!c[x].size()) {
            puts("-1") ;
            return 0 ;
        }
        j = upper_bound(c[x].begin() ,c[x].end(),k) - c[x].begin() ;
        if(j == c[x].end()-c[x].begin() ) {
            sum += c[x][0] + n - k ;
            k = c[x][0] ;
        }else {
            sum += c[x][j] - k ;
            k = c[x][j] ;
        }
    }
    cout << sum <<endl ; 
    return 0 ;
}