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 Ruby (2.3.3) (judge verdict: Accepted)

s = gets.chomp
t = gets.chomp

n = s.length

ss = Hash.new()

s = s + s
s.chars.each_with_index do |c, idx|
    ss[c] = Array.new unless ss[c]
    ss[c].push(idx)
end

ss.each do |k, v|
    ss[k].sort!
end

ans = 0
cur = n-1

t.chars.each do |c|
    unless ss[c]
        ans = -1
        break
    end

    r = ss[c].bsearch_index{|x| x > cur}
    nex = r ? ss[c][r] : ss[c].last

    nex %= n
    step = (nex - cur + n) % n
    step = n if step == 0

    ans += step
    cur = nex
end

puts ans