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 <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <stdio.h>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#define rep(i,a,n) for(ll i =(a);i<(n);++i)
#define urep(i,a,n) for(ll i = (a);i>=(n);--i)
#define all(x) (x).begin(),(x).end()
#define INF 1e18
typedef long long ll;
const ll MAX=510000;
const ll mod = 1e9 + 7;
using namespace std;
ll dx[4] = { 1,-1,0,0 };
ll dy[4] = { 0,0,1,-1 };
ll N,M,X,Y,A,B,C,D,Q,K,R,W,H,P,L,G;
ll ans;
string S,T;
ll h[101010];
ll y[101010];
ll a[201010];
ll k[101010];
ll b[201010];
ll d[101010];
ll t[101010];
ll p[101010];
ll n[101010];
ll l[101010];
ll r[101010];
ll v[101010];
ll s[101010];
ll x[101010];
ll gcd(ll a, ll b) {
	if (b == 0)return a;
	else return gcd(b, a%b);
}
ll lcm(ll a,ll b){
    return a/gcd(a,b)*b;
}
struct Edge{
	ll to,cost;
	 Edge(ll to, ll cost) : to(to), cost(cost) {}
};
typedef vector<vector<Edge> > AdjList;
AdjList graph;
vector<ll> dist;

struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
	vector<ll> size;
    UnionFind(ll N) : par(N),size(N) { //最初は全てが根であるとして初期化
        for(ll i = 0; i < N; i++) par[i] = i;
		for(ll i = 0; i < N; i++) size[i] = 1;
    }

    ll root(ll x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }
	ll GetSize(ll x){
		return size[root(x)];
	}

    void unite(ll x, ll y) { // xとyの木を併合
        ll rx = root(x); //xの根をrx
        ll ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
		size[ry]+=size[rx];

    }

    bool same(ll x, ll y) { // 2つのデータx, yが属する木が同じならtrueを返す
        ll rx = root(x);
        ll ry = root(y);
        return rx == ry;
    }
};

int main(){
    cin>>S>>T;
    map<char,ll> m1;
    map<char,ll> m2;
    ll num =0;
    rep(i,0,S.size()){
        m2[S[i]]+=1;
    }
    rep(i,0,T.size()){
        m1[T[i]]+=1;
        if(m2[T[i]]==0){
            cout<<-1<<endl;
            return 0;
        }
    }
    vector<vector<ll>> v(50);
    rep(i,'a','z'+1){
        rep(j,0,S.size()){
            if(S[j]==i){
                v[i-'a'].push_back(j+1);
            }
        }
    }
    ll ans =0;
    ll pos =0;
    rep(i,0,T.size()){
        if(upper_bound(v[T[i]-'a'].begin(),v[T[i]-'a'].end(),pos) == v[T[i]-'a'].end()){
            ans += S.size()-pos+*upper_bound(v[T[i]-'a'].begin(),v[T[i]-'a'].end(),0);
            pos = *upper_bound(v[T[i]-'a'].begin(),v[T[i]-'a'].end(),0);
        }else{
            ans += *upper_bound(v[T[i]-'a'].begin(),v[T[i]-'a'].end(),pos) - pos;    
            pos = *upper_bound(v[T[i]-'a'].begin(),v[T[i]-'a'].end(),pos);
        }
    }
    cout<<ans<<endl;
    return 0;
}