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

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

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 s of maximum possible length such that after removing this substring t will remain a subsequence of s.

If you want to remove the substring s[l;r] then the string s will be transformed to s_1 s_2 ... s_{l-1} s_{r+1} s_{r+2} ... s_{|s|-1} s_{|s|} (where |s| is the length of s).

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

Input

The first line of the input contains one string s consisting of at least 1 and at most 2 ⋅ 10^5 lowercase Latin letters.

The second line of the input contains one string t consisting of at least 1 and at most 2 ⋅ 10^5 lowercase Latin letters.

It is guaranteed that t is a subsequence of s.

Output

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

Examples

Input


bbaba
bb


Output


3


Input


baaba
ab


Output


2


Input


abcde
abcde


Output


0


Input


asdfasdf
fasd


Output


3
#include <bits/stdc++.h>
using namespace std;
int a[200005];
int main() {
  string s1, s2;
  cin >> s1 >> s2;
  int x = s1.size(), y = s2.size();
  while (y && x) {
    if (s1[x - 1] == s2[y - 1]) {
      a[y] = x;
      y--;
    }
    x--;
  }
  y = 1;
  a[s2.size() + 1] = s1.size() + 1;
  int res = a[1] - 1;
  for (int i = 0; i < s1.size(); ++i) {
    if (s1[i] == s2[y - 1]) {
      res = max(res, a[y + 1] - i - 2);
      y++;
    }
  }
  cout << res;
}
