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: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>

#include <set>
#include <map>
#include <iomanip>

using namespace std;

#define rep(i, n) for(long long i = 0; i < (long long)(n); i++)
#define all(x) (x).begin(),(x).end()

using ll = long long;

int main()
{
	const int M = 'z' - 'a' +1;
	vector<int> range(M, -1);

	string s, t;
	cin >> s >> t;

	int N = s.size();
	s.append(s);

	vector<vector<ll>> pos_mp(M);

	rep(i, s.size())
	{
		pos_mp[s[i] - 'a'].push_back(i);
	}

	ll ans = 0;
	rep(i, t.size())
	{
		int p = t[i] - 'a';

		if (pos_mp[p].size() == 0)
		{
			cout << -1 << endl;
			return 0;
		}
		
		ll now = ans % N;
		auto it = lower_bound(all(pos_mp[p]), now);
		ll next = *it;

		ans += next - now;
	}

	cout << ans + 1 << endl;

	return 0;
}