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Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.

Sergey gives Nastya the word t and wants to get the word p out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters‘ indices of the word t

: a1... a|t|. We denote the length of word x as |x|. Note that after removing one letter, the indices of other letters don‘t change. For example, if t?=?"nastya" and a?=?[4,?1,?5,?3,?2,?6] then removals make the following sequence of words "nastya" "nasya" "asya" "sya" "sa" "a" "".

Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word p

. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.

It is guaranteed that the word p can be obtained by removing the letters from word t.

Input

The first and second lines of the input contain the words t

and p, respectively. Words are composed of lowercase letters of the Latin alphabet (1?≤?|p|?<?|t|?≤?200?000). It is guaranteed that the word p can be obtained by removing the letters from word t.

Next line contains a permutation a1,?a2,?...,?a|t| of letter indices that specifies the order in which Nastya removes letters of t (1?≤?ai?≤?|t|, all ai are distinct).

Output

Print a single integer number, the maximum number of letters that Nastya can remove.

Examples input Copy

ababcba
abb
5 3 4 1 7 6 2

output Copy

3

input Copy

bbbabb
bb
1 6 3 4 2 5

output Copy

4

Note

In the first sample test sequence of removing made by Nastya looks like this:

"ababcba" "ababba" "abbba" "abba"

Nastya can not continue, because it is impossible to get word "abb" from word "ababcba".

So, Nastya will remove only three letters.

題目大意：給兩個字符串s1,s2和一個s1長度的數組od[ ]，問按od[]給定的順序去掉s1的字母，最多可以去掉多少個字母使s2仍然是s1的子串。

解題思路：本題暴力必定會超時，所以可以對數組進行二分查找。用一個Check函數檢查s1去掉前mid位是否還能使s2是s1的子串，可以則head = mid + 1，否則tail = mid - 1。

代碼：

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<string>
 6 #include<iostream>
 7 #include<algorithm>
 8 #include<map>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 typedef long long LL;
13 const LL MaxN = 2e5;
14 
15 string s1, s2;
16 int od[MaxN+5], len1, len2;
17 int head, tail, mid, cnt;
18 bool vis[MaxN + 5];
19 
20 int Check(int x)
21 {
22     memset(vis, 0, sizeof(vis));
23     cnt = 0;
24     for(int i = 0;i <= x;i++)
25     {
26         vis[od[i]-1] = 1;
27     }
28     for(int i = 0;i < len1;i++)
29     {
30         if(vis[i]) continue;
31         if(s1[i] == s2[cnt]) cnt++;
32         if(cnt == len2) return 1;
33     }
34     return 0;
35 }
36 
37 int main()
38 {
39     cin >> s1 >> s2;
40     len1 = s1.length();len2 = s2.length();
41     for(int i = 0;i < len1;i++)
42     {
43         scanf("%d", od + i);
44     }
45         head = 0, tail = len1-1;
46         while(head <= tail)
47         {
48             mid = (head + tail)/2;
49             if(Check(mid)) head = mid + 1;
50             else tail = mid - 1;
51             if(head > tail)
52             {
53                  if(tail == mid) printf("%d\n", mid+1);
54                  if(head == mid) printf("%d\n", mid);
55                  break;
56             }
57         }
58     return 0;
59 }

CodeForces - 779D String Game