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>
#pragma GCC optimize("O3")
using namespace std;
const long long INF = 1e12;
const int MOD = 1e9 + 7;
const unsigned long long BIT_FLAG_0 = (1 << 0);
const unsigned long long BIT_FLAG_1 = (1 << 1);
const unsigned long long BIT_FLAG_2 = (1 << 2);
const unsigned long long BIT_FLAG_3 = (1 << 3);
const unsigned long long BIT_FLAG_4 = (1 << 4);
const unsigned long long BIT_FLAG_5 = (1 << 5);
const unsigned long long BIT_FLAG_6 = (1 << 6);
const unsigned long long BIT_FLAG_7 = (1 << 7);
const long long dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const long long Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1},
                Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
void print() { cout << "\n"; }
template <class Head, class... Tail>
void print(Head &&head, Tail &&...tail) {
  cout << head;
  if (sizeof...(tail) != 0) cout << " ";
  print(forward<Tail>(tail)...);
}
template <class T>
void print(vector<T> &vec) {
  for (auto &a : vec) {
    cout << a;
    if (&a != &vec.back()) cout << " ";
  }
  cout << "\n";
}
template <class T>
void print(set<T> &set) {
  for (auto &a : set) {
    cout << a << " ";
  }
  cout << "\n";
}
template <class T>
void print(vector<vector<T>> &df) {
  for (auto &vec : df) {
    print(vec);
  }
}
long long power(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a;
    a *= a;
    n >>= 1;
  }
  return res;
}
long long comb(long long n, long long k) {
  vector<vector<long long>> v(n + 1, vector<long long>(n + 1, 0));
  for (long long i = 0; i < v.size(); i++) {
    v[i][0] = 1;
    v[i][i] = 1;
  }
  for (long long k = 1; k < v.size(); k++) {
    for (long long j = 1; j < k; j++) {
      v[k][j] = (v[k - 1][j - 1] + v[k - 1][j]);
    }
  }
  return v[n][k];
}
void add(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
__attribute__((constructor)) void faster_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
string s, t;
signed main() {
  cin >> s >> t;
  long long slen = (long long)((s).size());
  long long tlen = (long long)((t).size());
  vector<long long> rightmost(tlen, -1);
  vector<long long> leftmost(tlen, -1);
  {
    long long cur = 0;
    for (long long ni = 0; ni < (long long)(slen); ni++) {
      if (cur == tlen) {
        break;
      }
      if (s[ni] == t[cur]) {
        leftmost[cur] = ni;
        cur++;
      }
    }
  }
  {
    long long cur = tlen - 1;
    for (long long ni = (long long)(slen)-1; ni >= 0; ni--) {
      if (cur == -1) {
        break;
      }
      if (s[ni] == t[cur]) {
        rightmost[cur] = ni;
        cur--;
      }
    }
  }
  long long ret = 0;
  for (long long ti = 0; ti < (long long)(tlen - 1); ti++) {
    long long dis = rightmost[ti + 1] - leftmost[ti] - 1;
    chmax(ret, dis);
  }
  chmax(ret, rightmost[0]);
  chmax(ret, slen - leftmost[tlen - 1] - 1);
  print(ret);
  return 0;
}
