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 <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <bitset>
#include<vector>
#include<utility>
#include<fstream>
#include<queue>
#include <iomanip>
using namespace std;
typedef long long ll;
#define rep(i,k) for(i=0;i<k;i++)
#define P pair<ll,ll>
ll n,p,k, b, c, i, j, x, y, z = 0, m = 0,q;
ll a[100] = {}, yy[100] = {}, aa[100] = {};
double xx,zz=0;
string s,t;
P pp[200005];
vector<ll> v[100];
char ss[100000], tt[100000];
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin>>s>>t ;
	b = s.size();
	c = t.size();


	rep(i, b) {
		x = int(char(s[i])) - int('a');
		v[x].push_back(i+1);
		aa[x]++;
	}
	rep(i, c) {
		x = int(char(t[i])) - int('a');
		if (aa[x] == 0) {
			cout << "-1";
			return 0;
		}
		else {
			rep(j, aa[x]) {
				if (m < v[x][j]) {
					m = v[x][j];
					break;
				}

					m = v[x][0];
					z += b;
				}

		}
	}
	z += m;
	
	cout <<z;
	return 0;
}