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<bits/stdc++.h>
using namespace std;
string s,t;
int dp[100009],num[100009],tmp[26];
vector<int>pos[26];
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>s>>t;
	for(int i=0;i<s.size();i++) pos[s[i]-'a'].push_back(i);
	for(int i=0;i<t.size();i++)
		if(pos[t[i]-'a'].size()==0) {
			cout<<-1<<endl;
			return 0;
		}
		else num[i]=pos[t[i]-'a'][min(tmp[t[i]-'a']++,(int)pos[t[i]-'a'].size()-1)];
	int len=1;
	dp[1]=-num[0];
	for(int i=1;i<t.size();i++)
		if(-num[i]>=dp[len]) dp[++len]=-num[i];
		else dp[upper_bound(dp+1,dp+len,-num[i])-dp]=-num[i];
	printf("%lld\n",1ll*s.size()*(len-1)+1ll*num[t.size()-1]+1ll*1);
	return 0;
}