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 <type_traits>
using namespace std;

using ll = long long;

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define ddrep(i, n) for (int i = n; i > 0; --i)
#define srep(i, s, t) for (int i = s; i < t; ++i)
#define ssrep(i, s, t) for (int i = s; i <= t; ++i)

#define rng(a) a.begin(), a.end()
#define rrng(a) a.rbegin(), a.rend()
#define pb push_back
#define eb emplace_back
#define fi first
#define se second

#define chmax(x, y) (x = max(x, y))
#define chmin(x, y) (x = min(x, y))

using pi = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vs = vector<string>;
using vvs = vector<vs>;
using ld = long double;

template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
    os << "(" << p.first << "," << p.second << ")";
    return os;
}

template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
    os << "{";
    rep(i, (int)v.size()) {
        if (i)
            os << ",";
        os << v[i];
    }
    os << "}";
    return os;
}

template <typename T, size_t S> void printArray(const T (&array)[S]) {
    for (auto val : array)
        std::cout << val << ", ";
    std::cout << "\n";
}

const int mod = 1e9 + 7;
const int inf = 1e9 + 5;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << std::setprecision(10);

    string s, t; std::cin >> s >> t;
    vvi idx(27);
    int slen = s.size(), tlen = t.size();
    rep(i, slen) {
        idx[s[i]-'a'].pb(i);
    }
    ll cur = -1, ans = 0;
    rep(i, tlen) {
        if (idx[t[i]-'a'].size() == 0) {
            ans = -1;
            break;
        }
        auto it = upper_bound(rng(idx[t[i]-'a']), cur);
        if (it == idx[t[i]-'a'].end()) {
            ans += slen-(cur+1);
            ans += idx[t[i]-'a'][0]+1;
            cur = idx[t[i]-'a'][0];
        } else {
            ans += *it-cur;
            cur = *it;
        }
    }
    std::cout << ans << "\n";
}