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>
using namespace std;
#define next Next
/*char buf[1<<21],*p1=buf,*p2=buf;
inline int gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}*/
#define gc getchar
inline int read()
{
	int ret=0,f=0;char c=gc();
	while(!isdigit(c)){if(c=='-')f=1;c=gc();}
	while(isdigit(c)){ret=ret*10+c-48;c=gc();}
	if(f)return -ret;return ret;
}
#define N 100005
char a[N << 1], b[N];
int n, m, nx[N << 1][26];
int main(){
	scanf("%s%s", a + 1, b + 1);
	n = strlen(a + 1), m = strlen(b + 1);
	for (register int i = 1; i <= n; ++i) a[n + i] = a[i];
	for (register int i = 0; i < 26; ++i) nx[n << 1][i] = -1;
	for (register int i = (n << 1) - 1; ~i; --i){
		for (register int j = 0; j < 26; ++j) nx[i][j] = nx[i + 1][j];
		nx[i][a[i + 1] - 'a'] = i % n + 1;
	}
	int k = 0, s = 0;
	for (register int i = 1; i <= m; ++i)
		if (~nx[k][b[i] - 'a']){
			if (nx[k][b[i] - 'a'] <= k) ++s;
			k = nx[k][b[i] - 'a'];
		}
		else return printf("-1\n"), 0;
	printf("%lld\n", 1ll * s * n + k);
}