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)

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

#ifdef LOCAL
#include "../lib/dump.hpp"
#else
#define dump(...)
#define dumpv(...)
#endif

typedef long long ll;
typedef pair<int, int> P;
const int MOD = 1e9 + 7;
const int INF = 1001001001;
const ll LINF = 1001002003004005006ll;

void solve()
{
  string s, t;
  cin >> s;
  cin >> t;
  vector<vector<int>> p(26);
  for (int i = 0; i < s.size(); i++) {
    p[s[i] - 'a'].push_back(i);
  }
  dump(p);
  ll ans = 0;
  ll loop = 0;
  vector<int> c(26);
  for (int i = 0; i < t.size(); i++) {
    int j = t[i] - 'a';
    if (p[j].size() == 0) {
      // 見つからない
      cout << -1 << endl;
      return;
    } else {
      // 見つかった
      if (ans <= p[j][c[j]]) {
        // 先にある
        ans = p[j][c[j]] + 1;
        c[j] = (c[j] + 1) % p[j].size();
      } else {
        // 手前にある
        ans = p[j][c[j]] + 1;
        c[j] = (c[j] + 1) % p[j].size();
        loop++;
      }
      dump(t[i], ans, loop);
    }
  }
  cout << loop * s.size() + ans << endl;
}

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  // freopen("temp.1", "r", stdin);
  solve();
  return 0;
}