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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace __gnu_pbds;
using namespace std;
 
typedef tree<int, null_type, less<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;
 
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define PSET(x,y) fixed<<setprecision(y)<<x
#define pb push_back
#define pf push_front
#define pii pair<int,int>
#define mp make_pair
#define vi vector<int>
#define vip vector<pair<int,int> >
#define ff first
#define ss second
#define int long long
#define SIZE 200010
#define mod 1000000007
#define BIG 998244353
#define s(t) scanf("%d",&t)
#define p(t) printf("%d\n",t)
#define mii map<int,int>
#define MSET(table,i) memset(table, i, sizeof(table))
#define endl '\n'
#define tc int t;cin>>t;while(t--)
#define pi  3.1415926358
#define print(x) for(auto el: x) cout<<el<<" ";cout<<endl
#define bits(x) __builtin_popcount(x)
#define minimum(a,n) min_element(a,a+n)-a
#define maximum(a,n) max_element(a,a+n)-a
#define pqbig priority_queue<int>
#define pqsmall priority_queue<int,vector<int>,greater<int> >
#define all(v) v.begin(),v.end()
#define sitr set<int>:: iterator
#define mitr map<int,int>:: iterator
 
#define trace1(x)                cerr<<#x<<": "<<x<<endl
#define trace2(x, y)             cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)          cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d)       cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define trace5(a, b, c, d, e)    cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define trace6(a, b, c, d, e, f) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
 
const int N=30;
vi adj[N];

int32_t main()
{
    fast;
    string s,t;cin>>s>>t;
    int n=s.size();
    int m=t.size();
    for(int i=0;i<n;i++)
        adj[s[i]-'a'].pb(i);

    for(int i=0;i<m;i++)
    {
        if(adj[t[i]-'a'].size()==0)
        {
            cout<<-1;
            return 0;
        }
    }

    int complete_rounds=0;
    int prev=-1;
    for(int i=0;i<m;i++)
    {
        int val=t[i]-'a';
        int id=lower_bound(all(adj[val]),prev)-adj[val].begin();
        if(id==adj[val].size())
        {
            complete_rounds++;
            prev=adj[val][0];
        }
        else
            prev=adj[val][id];
    }

    cout<<n*complete_rounds+prev+1;

}