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;

char s[100005], t[100005];
int suf[100005][26];
long long ans;

int main(void) {
  scanf("%s%s", s + 1, t + 1);
  int ns = strlen(s + 1), nt = strlen(t + 1);
  for (int i = ns; i >= 1; --i) {
    memcpy(suf[i - 1], suf[i], sizeof suf[i]);
    suf[i - 1][s[i] - 'a'] = i;
  }
  int cur = 0;
  for (int i = 1; i <= nt; ++i) {
    int nxt = suf[cur][t[i] - 'a'];
    if (nxt == 0)
      nxt = suf[0][t[i] - 'a'];
    if (nxt == 0) {
      printf("-1\n");
      return 0;
    }
    ans += (nxt - cur + ns) % ns;
    if (nxt == cur)
      ans += ns;
    cur = nxt;
  }
  printf("%lld\n", ans);
  return 0;
}