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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 1E5;
int nextOcc[26][MAXN];

int main()
{
    string s, t;
    cin >> s >> t;
    int n = (int) s.length();
    for (int i = 0; i < 26; i++)
    {
        if (s[n - 1] == 'a' + i)
        {
            nextOcc[i][n - 1] = n - 1;
        }
        else
        {
            nextOcc[i][n - 1] = -1;
        }
    }
    for (int i = 0; i < 26; i++)
    {
        for (int j = n - 2; j >= 0; j--)
        {
            if (s[j] == 'a' + i)
            {
                nextOcc[i][j] = j;
            }
            else
            {
                nextOcc[i][j] = nextOcc[i][j + 1];
            }
        }
    }
    int index = 0;
    long long ans = 0;
    for (int i = 0; i < (int) t.length(); i++)
    {
        int key = t[i] - 'a';
        if (nextOcc[key][index] != -1)
        {
            ans += nextOcc[key][index] - index + 1;
            index = (nextOcc[key][index] + 1) % n;
        }
        else if (nextOcc[key][0] != -1)
        {
            ans += n - index;
            index = (nextOcc[key][0] + 1) % n;
            ans += nextOcc[key][0] + 1;
        }
        else
        {
            ans = -1;
            break;
        }
    }
    cout << ans << endl;
    return 0;
}