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)

def e_strings_of_inpurity():
    from bisect import bisect_left
    S = input()
    T = input()

    char_appear = {ch: [] for ch in 'abcdefghijklmnopqrstuvwxyz'}
    for i, ch in enumerate(S):
        char_appear[ch].extend([i, i + len(S)])  # Sを2個連結させた文字列を考える

    ans = 0
    # editorial における二分探索を使うやり方
    p = 0  # 文字 t は S' の p-1 (mod |S|) 番目に現れた
    for t in T:
        if len(char_appear[t]) == 0:
            return -1  # 文字tが存在しなかった
        p = char_appear[t][bisect_left(char_appear[t], p)] + 1
        if p >= len(S):
            p %= len(S)
            ans += len(S)
    ans += p  # 残った分を足す
    return ans

print(e_strings_of_inpurity())