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 Python (3.4.3) (judge verdict: Wrong Answer)

import bisect

def ctoi(char):
  return ord(char) - 97

alphabets = [[] for _ in range(26)]

#素材
ingredients = input()
#つくりたいもの
recipe = input()

# 素材のアルファベット分布リストをつくる
for i in range(len(ingredients)):
  alpha_idx = ctoi(ingredients[i])
  alphabets[alpha_idx].append(i)

ingredients_counter = [0] * 26
for i in range(26):
  ingredients_counter[i] = len(alphabets[i])
ingredients_pointer = -1
ingredients_idx_list = []
flag = True
for i in range(len(recipe)):
  alpha_idx = ctoi(recipe[i])
  if ingredients_counter[alpha_idx] == 0:
    flag = False
    break
  #今のingredients_pointerより右側に使える素材があるか探す
  pointer_candidate_idx = \
  	bisect.bisect_right(alphabets[alpha_idx], ingredients_pointer)
  #右側に無かった。ループして先頭のを使ってどうぞ
  if pointer_candidate_idx == ingredients_counter[alpha_idx]:
    ingredients_pointer = alphabets[alpha_idx][0]
  else:
    ingredients_pointer = alphabets[alpha_idx][pointer_candidate_idx]
  ingredients_idx_list.append(ingredients_pointer)

if flag == False:
  print(-1)
else: 
  loop = 0
  idx = ingredients_idx_list[0]
  #idxのリセット分ループを数える。左より右が小さかったらループ
  for i in range(1, len(ingredients_idx_list)):
    if ingredients_idx_list[i] < idx:
      loop += 1
    idx = ingredients_idx_list[i]
  # ループx素材の長さ + つくりたいものの最後の文字のインデックス(に、+１)
  print(loop * len(ingredients) + ingredients_idx_list[-1] + 1)