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)

/*input
contest
son


*/
 
/*
                                                    ______________
                                                          |       )
                                                          |   )   /
                                                          |______/
                                                    |     |      \
                                                    |     |   )   \
                                                    |_____|_______)
 
*/
 
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <iomanip>
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll , ll> PLL;
typedef long double ld;
 
#define pb push_back
#define all(c) c.begin(),c.end()
#define allr(c) c.rbegin(),c.rend()
int mod = 1000000007;
const int inf = 1034567891;
const ll LL_INF = 1234567890123456789ll;
#define PI 3.14159265
// #define endl '\n'
#define F first
#define S second
#define debug(x) cout << #x << " = " << x << endl;
#define TRACE
 
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
 
#define out(container) for (auto it : container) cout << it << " "; cout << endl;
 
 
template < typename T > T GCD(T a, T b)            { ll t; while(a) { t = a; a = b % a; b = t; } return b; }
template < typename T > string toString(T a)       { return to_string(a); }
template < typename T > void toInt(string s, T &x) { stringstream str(s); str >> x;}
inline int add(int x, int y){ x += y; if(x >= mod) x -= mod; return x;}
inline int sub(int x, int y){ x -= y; if(x < 0) x += mod; return x;}
inline int mul(int x, int y){ return (x * 1ll * y) % mod;}
inline int powr(int a, ll b){
  int x = 1 % mod;
  while(b){
    if(b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a){ return powr(a, mod - 2);}

int cnt1[26], cnt2[26];
vector <int> vec[26];

int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  string str1, str2;
  cin >> str1 >> str2;
  int n = str1.size();
  int m = str2.size();
  for (int i = 0; i < n; i++) {
    auto it = str1[i];
    cnt1[it - 97]++;
    vec[it - 97].pb(i);
  }
  for (int i = 0; i < m; i++) {
    auto it = str2[i];
    cnt2[it - 97]++;
  }
  for (int i = 0; i < 26; i++) {
    if (cnt2[i] && !cnt1[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  int cur = -1;
  ll ans = n;
  bool ok = false;
  for (int i = 0; i < m; i++) {
    int v = str2[i] - 97;
    int ind = lower_bound(all(vec[v]), cur) - vec[v].begin();
    if (ind == vec[v].size()) {
      cur = -1;
      ind = lower_bound(all(vec[v]), cur) - vec[v].begin();
      ans += n;
      ok = true;
    }
    cur = vec[v][ind];
    if (i == m - 1) {
      ans -= n;
      ans += cur + 1;
    }
  }
  if (!ok) {
    ans -= n;
    ans += cur + 1;
  }
  cout << ans << endl;


  return 0;
}