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 OCaml (4.10.0) (judge verdict: Accepted)

let ioc c = int_of_char c - int_of_char 'a'

let s = read_line()
let t = read_line()
let n = String.length s
let m = String.length t
let d = Array.init 26 (fun _ -> Array.make (n + 1) ~-1)

let rec f i j s =
  if i = m then s + j
  else
    let k = ioc t.[i] in
    if d.(k).(0) = -1 then -1
    else
      let j = d.(k).(j) in
      if j = -1 then f (i + 1) (d.(k).(0) + 1) (s + n)
      else f (i + 1) (j + 1) s

let () =
  for i = 0 to n - 1 do
    d.(ioc s.[i]).(i) <- i
  done;

  Array.iter (fun d ->
    for i = 0 to n - 1 do
      let j = n - 1 - i in
      if d.(j) = -1 then
        d.(j) <- d.(j + 1)
      done
  ) d;

  d.(ioc s.[0]).(0) <- 0;

  Printf.printf "%d\n" (f 0 0 0)