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: Wrong Answer)

require 'time'

def calcs(x,y,ax,ay,bx,by) ((ax-x)*(by-y)-(bx-x)*(ay-y))/2.0 end

def inpf() a=gets.chomp.split(" ").map(&:to_f)end

def inps() a=gets.chomp.split(" ")end

def copy(a) Marshal.load(Marshal.dump(a)) end

def factorial(n)(n < 2)? 1 : (2..n).inject(:*) end

def scount(a) b=na(a.max+1);a.each{|n|b[n]+=1};return b end

def na(n,d=0) Array.new(n,d)end

def na2(n,m,d=0) Array.new(n){Array.new(m,d)}end

def na3(n,m,l,d=0) Array.new(n){Array.new(m){Array.new(l,d)}}end

def yn(f) return (f)? "Yes": "No" end

def ynb(f) return (f)? "YES": "NO" end

def ap(a) a.each{|u| p u} end

def bit(n) n.to_s(2).split("").map(&:to_i) end

def inp() a=gets.chomp.split(" ").map(&:to_i)end

s = inps[0]
t = inps[0]
dp = na2(26,s.size,nil)
l = s.size
(1..(s.size)).each do |i|
  26.times do |j|
    dp[j][l-i] = dp[j][l-i+1]
  end
  pp = s[l-i].ord - "a".ord
  dp[pp][l-i] = l-i
end
now_index = 0
rc =0
t.each_char do |c|
  cd = c.ord - "a".ord
  if(dp[cd][now_index] == nil)
    if(dp[cd][0] == nil)
      p -1
      exit
    else
      rc += 1
      now_index = dp[cd][0]
    end
  else
    now_index = dp[cd][now_index]
  end
end
p rc * s.size + now_index +1
#puts ()? "Yes" : "No"
#puts ()? "YES" : "NO"
#puts ()? "yes" : "no"