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>
#include<time.h>
#include<stdlib.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse4")
#define pb push_back
#define mp make_pair
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long
#define pll pair<ll,ll>
#define vll vector<ll,ll>
#define all(a) (a).begin(),(a).end()
#define ff first
#define ss second
#define sz(x) (ll)x.size()
#define hell 1000000007
#define repf(i,a,b) for(ll i=a;i<b;i++)
#define repb(i,a,b) for(ll i=a;i>b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define gcd(a,b) __gcd((a),(b))
#define EPS 1e-12
#define lcm(a,b) ((a)*(b))/__gcd((a),(b))
#define sqrt sqrtl
#define cbrt cbrtl
using namespace std;
string s,t;
vector<ll> pos[30];
ll cnt[30];
int main()
{
  IOS
  cin>>s>>t;
  ll i;
  
  for(i=0;i<30;i++)
  	pos[i].clear();
  ll l=s.length();
  for(i=0;i<l;i++)
  	{
  		cnt[s[i]-'a']=1;
  		pos[s[i]-'a'].pb(i);
  	}
  ll l1=t.length();
  for(i=0;i<l1;i++)
  	if(!cnt[t[i]-'a'])
  	{
  		cout<<-1;
  		return 0;
  	}
  ll prev=-1;
  ll sum=0;
  for(i=0;i<l1;i++)
  {
  	 ll ub=upper_bound(all(pos[t[i]-'a']),prev)-pos[t[i]-'a'].begin();
  	 if(ub==pos[t[i]-'a'].size())
  	 {
  	 	sum+=l;
  	 	prev=-1;
  	 	ll ub=upper_bound(all(pos[t[i]-'a']),prev)-pos[t[i]-'a'].begin();
  	 	prev=pos[t[i]-'a'][ub];
  	 }
  	 else
  	 	prev=pos[t[i]-'a'][ub];
  }
  sum+=prev+1;
  cout<<sum;
  return 0;
}