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
//------------------------------------------
#include <bits/stdc++.h>
using namespace std;

//typedef
//------------------------------------------
using LL = int64_t;
using VL = vector<LL>;
using VVL = vector<VL>;
using VS = vector<string>;
using PLL = pair<LL, LL>;

//container util
//------------------------------------------
#define whole(f,x,...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x)
#define rwhole(f,x,...) ([&](decltype((x)) whole) { return (f)(rbegin(whole), rend(whole), ## __VA_ARGS__); })(x)
#define EACH(i,c) for(decltype((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) whole(sort, c)
#define RSORT(c) rwhole(sort, c)

//constant
//--------------------------------------------
constexpr double EPS = 1e-10;
constexpr double PI  = M_PI;
constexpr int MOD = 1000000007;

// grid
//--------------------------------------------
VL dx = {0, 1, 0, -1};
VL dy = {1, 0, -1, 0};
VL dx2 = {-1, 0, 1, -1, 1, -1, 0, 1};
VL dy2 = {-1, -1, -1, 0, 0, 1, 1, 1};

//debug
//--------------------------------------------
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

//IO accelerate
//--------------------------------------------
struct InitIO {
    InitIO() {
        cin.tie(nullptr);
        ios_base::sync_with_stdio(false);
        cout << fixed << setprecision(30);
    }
} init_io;

//template
//--------------------------------------------
// declaretion
template<typename T> istream& operator >>(istream& is, vector<T>& vec);
template<typename T1, typename T2> ostream& operator <<(ostream& os, const pair<T1, T2>& p);
template<typename T> ostream& operator <<(ostream& os, const vector<T>& vec);
template<typename T> ostream& operator <<(ostream& os, const vector<vector<T>>& vv);
template<typename T> vector<T> make_v(size_t a);
template<typename T,typename... Ts> auto make_v(size_t a,Ts... ts);
template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type fill_v(T &t,const V &v);
template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type fill_v(T &t,const V &v);

// implementation
template<typename T>
istream& operator >>(istream& is, vector<T>& vec) {
	for(T& x: vec) is >> x;
	return is;
}
template<typename T1, typename T2>
ostream& operator <<(ostream& os, const pair<T1, T2>& p) {
	os << p.first << "," << p.second;
	return os;
}
template<typename T>
ostream& operator <<(ostream& os, const vector<T>& vec) {
	for(int i=0; i<vec.size(); i++){
		os << vec[i] << ( i+1 == vec.size() ? "" : "\t" );
	}
	return os;
}
template<typename T>
ostream& operator <<(ostream& s, const vector<vector<T>>& vv) {
	for (int i = 0; i < vv.size(); ++i) {
		s << vv[i] << endl;
	}
	return s;
}

// 多重vector
// auto dp=make_v<int>(4,h,w) みたいに使える
template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}

template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
	return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}

// 多重vectorのためのfill
// fill_v(dp,0) みたいに使える
template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}

template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){
	for(auto &e:t) fill_v(e,v);
}

template<class T,class U> void chmax(T&a,U b){if(a<b)a=b;}
template<class T,class U> void chmin(T&a,U b){if(b<a)a=b;}
template<typename T> T gcd(T a, T b) { return b?gcd(b,a%b):a;}
template<typename T> T lcm(T a, T b) { return a/gcd(a,b)*b;}

//main code
int main(int argc, char *argv[])
{
	string s,t;
	cin >> s >> t;
	LL n = s.size();
	map<char, VL> mp;
	for (int i = 0; i < n; i++) {
		mp[s[i]].emplace_back(i);
	}
	LL loop = 0;
	LL now = -1;
	for (int i = 0; i < t.size(); i++) {
		if (mp[t[i]].empty()) {
			cout << -1 << endl;
			return 0;
		}
		if (mp[t[i]].back() <= now) {
			loop++;
			now = mp[t[i]].front();
		} else {
			now = *upper_bound(mp[t[i]].begin(), mp[t[i]].end(), now);
		}
	}
	LL ans = n*loop + now+1;
	cout << ans << endl;
	return 0;
}