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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <queue>

using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<vector<int> > vvint;
typedef vector<long long> vll, vLL;
typedef vector<vector<long long> > vvll, vvLL;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<n;++i)
#define mod (ll)(1e9+7)
#define FIX(a) ((a)%mod+mod)%mod
#define ALL(obj) (obj).begin(), (obj).end()
#define rALL(obj) (obj).rbegin(), (obj).rend()
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define fi first
#define se second
#define pb push_back
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    string s,t;
    cin >> s >> t;
    vint ss(26),ts(26);
    int ls = s.length();
    int lt = t.length();
    REP(i,ls){
        ss[(s[i]-'a')]+=1;
    }
    REP(i,lt){
        ts[(t[i]-'a')]+=1;
    }
    REP(i,26){
        if(ts[i]>0&&ss[i]==0){
            cout <<"-1\n";
            return 0;
        }
    }
    vvint sx(26);
    REP(i,ls){
        sx[(s[i]-'a')].pb(i);
    }
    REP(i,26){
        sort(ALL(sx[i]));
    }
    ll ans = 0;
    int now = -1;
    REP(i,lt){
        //cout << i << endl;
        char c = t[i];
        ll ok = (ll)(sx[(c-'a')].size());
        ll ng = -1;
        while(abs(ok-ng)>1){
            ll mid = (ok+ng)/2;
            if(sx[(c-'a')][mid]>now){
                ok = mid;
            }else{
                ng = mid;
            }
        }
        if(ok==(int)(sx[(c-'a')].size())){
            ans++;
            now=-1;
            ng=-1;
            while(abs(ok-ng)>1){
                ll mid = (ok+ng)/2;
                if(sx[(c-'a')][mid]>now){
                    ok = mid;
                }else{
                    ng = mid;
                }
            }
            now=sx[(c-'a')][ok];
        }else{
            now=sx[(c-'a')][ok];
        }
        //cout <<ans << " "<< c << " " << now << endl;
    }
    cout << (ll)ans*(ll)ls+(ll)now+1 << endl;;
    return 0;
}