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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
int main()
{
	char s[100005], t[100005];
	scanf("%s%s", s, t);
	long long int i, n, m;
	n = strlen(s);
  	m = strlen(t);
	long long int count[30];
	for (i = 0; i < 30; i++)
		count[i] = 0;
	long long int p[30][100005];
	for (i = 0; i < n; i++)
	{
		p[s[i] - 'a'][count[s[i] - 'a']] = i;
		count[s[i] - 'a']++;
	}
	long long int c[30];
	for (i = 0; i < 30; i++)
		c[i] = 0;
	for (i = 0; i < m; i++)
		c[t[i] - 'a']++;
	for (i = 0; i < 30; i++)
	{
		if (count[i] == 0 && c[i] > 0)
		{
			printf("-1\n");
			return 0;
		}
	}
	long long int ans = 1;
	long long int now = -1;
	long long int min, max, mid;
	for (i = 0; i < m; i++)
	{
		if (now < p[t[i] - 'a'][0])
			now = p[t[i] - 'a'][0];
		else if (now >= p[t[i] - 'a'][count[t[i] - 'a']])
		{
			ans++;
			now = p[t[i] - 'a'][0];
		}
		else
		{
			min = 0;
			max = count[t[i] - 'a'] - 1;
			while (max - min > 1)
			{
				mid = (max + min) / 2;
				if (p[t[i] - 'a'][mid] <= now)
					min = mid;
				else
					max = mid;
			}
			now = p[t[i] - 'a'][max];
		}
	}
  	ans--;
  	now++;
	printf("%lld\n", ans * n + now);
	return 0;
}