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)

// https://atcoder.jp/contests/abc138/tasks/abc138_e
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
#define EPS (1e-9)
#define INF (1e9)
#define INFL (1e18)
#define MOD (1000000007)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
#define BIT(n) (1LL << (n))
// ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
// ll lcm(ll c, ll d) { return c / gcd(c, d) * d; }

int dp[100005];
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    string s, t;
    multimap<char, int> mp;
    cin >> s >> t;

    for (int i = 0; i < (int)s.size(); i++) {
        mp.insert({s[i], i});
    }
    set<char> s1, s2;
    for (auto x : s) {
        s1.insert(x);
    }
    for (auto x : t) {
        s2.insert(x);
    }
    for (auto x : s2) {
        if (s1.count(x) == 0) {
            cout << "-1\n";
            return 0;
        }
    }
    int pos = 0;
    ll ans = 1;
    int len = s.length();
    int i = 0;
    while (i < (int)t.length()) {
        auto ret = mp.equal_range(t[i]);
        bool find = false;
        for (auto it = ret.first; it != ret.second; ++it) {
            if (pos <= it->second) {
                ans += it->second - pos;
                pos = it->second;
                find = true;
                break;
            }
        }
        if (find == false) {
            ans += len - pos;
            pos = 0;
        } else {
            i++;
        }
    }
    cout << ans << endl;

    return 0;
}