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)

s = input()
t = input()
 
flg = False
for x in t:
  if not x in s:
    flg = True
  
if flg:
  print(-1)
else:
  dic = {}
  for i in range(len(s)):
    dic[s[i]] = dic.get(s[i], []) + [i]
  for i in range(len(s)):
    dic[s[i]] = dic.get(s[i], []) + [i+len(s)]

  ans = 0
  p = 0
  for i in range(len(t)):
    if not t[i] in dic:
      print(-1)
      break
    else:
      idx = [-1]+dic[t[i]]
      l, r = 1, len(idx)
      while l<=r:
        m = (l+r)//2
        if p>idx[m-1] and p<=idx[m]:
          p = idx[m]+1
          break
        elif p<idx[m-1]:
          r = m-1
        else:
          l = m+1
      if p>=len(s):
        p -= len(s)
        ans += len(s)
  ans += p
  print(ans)