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)

def ok?(s, t)
  hs = Hash.new(0)
  s.chars.map { |c| hs[c] += 1 }
  t.chars.map { |c| return if hs[c] == 0 }
  true
end

s = gets.chomp
t = gets.chomp

if !ok?(s, t)
  p -1
  exit
end

pos = Hash.new{|h,k| h[k]=[]}
s.each_char.with_index do |c, i|
  pos[c] << i
end
l = 0
prev_i = pos[t[0]][0]
t[1..-1].each_char do |c|
  j = pos[c].bsearch{|x| x > prev_i}
  prev_i = j || pos[c][0]
  l += 1 if !j
end

p s.size * l + (prev_i + 1)