D2. Remove the Substring (hard version)（思维 ）

D2. Remove the Substring (hard version)

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The only difference between easy and hard versions is the length of the string.

You are given a string ss and a string tt, both consisting only of lowercase Latin letters. It is guaranteed that tt can be obtained from ss by removing some (possibly, zero) number of characters (not necessary contiguous) from ss without changing order of remaining characters (in other words, it is guaranteed that tt is a subsequence of ss).

For example, the strings "test", "tst", "tt", "et" and "" are subsequences of the string "test". But the strings "tset", "se", "contest" are not subsequences of the string "test".

You want to remove some substring (contiguous subsequence) from ss of maximum possible length such that after removing this substring tt will remain a subsequence of ss.

If you want to remove the substring s[l;r]s[l;r] then the string ss will be transformed to s1s2…sl−1sr+1sr+2…s|s|−1s|s|s1s2…sl−1sr+1sr+2…s|s|−1s|s| (where |s||s| is the length of ss).

Your task is to find the maximum possible length of the substring you can remove so that tt is still a subsequence of ss.

Input

The first line of the input contains one string ss consisting of at least 11 and at most 2⋅1052⋅105 lowercase Latin letters.

The first line of the input contains one string tt consisting of at least 11 and at most 2⋅1052⋅105 lowercase Latin letters.

It is guaranteed that tt is a subsequence of ss.

Output

Print one integer — the maximum possible length of the substring you can remove so that tt is still a subsequence of ss.

Examples

input

bbaba
bb

output

3

input

baaba
ab

output

2

input

abcde
abcde

output

0

input

asdfasdf
fasd

output

3

算法：思维

 题解：匹配字串问题，我们可以先找到第一个子串出现的位置，记录到一个数组f里面，然后从后往前匹配，每次获取最大的长度，前面那个字符第一次出现的位置，如果a串中有一个字符和b串的字符相等，就向前推一个字符，继续寻找。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
 
using namespace std;
 
const int maxn = 2e5+7;
 
typedef long long ll;

string a;
string b;
int f[maxn];
 
int main() {
    cin >> a >> b;
    f[0] = -1;
    int k = 1;
    for(int i = 0, j = 0; i < a.size(); i++) {
        if(j < b.size() && a[i] == b[j]) {      //找到第一个子串的位置
            f[k++] = i;
            j++;
        }
    }
    int ans = 0;
    for(int i = a.size() - 1, j = 0; i >= 0; i--) {
        ans = max(ans, i - f[b.size() - j]);
        if(j < b.size() && a[i] == b[b.size() - j - 1]) {
            j++;
        }
    }
    cout << ans << endl;
    return 0;
}