description: Mr. Chanek gives you a sequence a indexed from 1 to n. Define f(a) as the number of indices where a_i = i. 
  
  You can pick an element from the current sequence and remove it, then concatenate the remaining elements together. For example, if you remove the 3-rd element from the sequence [4, 2, 3, 1], the resulting sequence will be [4, 2, 1]. 
  
  You want to remove some elements from a in order to maximize f(a), using zero or more operations. Find the largest possible f(a).
  
  Input
  
  The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the initial length of the sequence.
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2 ⋅ 10^5) — the initial sequence a.
  
  Output
  
  Output an integer denoting the largest f(a) that can be obtained by doing zero or more operations.
  
  Examples
  
  Input
  
  
  7
  2 1 4 2 5 3 7
  
  
  Output
  
  
  3
  
  
  Input
  
  
  4
  4 2 3 1
  
  
  Output
  
  
  2
  
  Note
  
  In the first example, f(A) = 3 by doing the following operations.
  
  [2,1,4,2,5,3,7] → [2,1,2,5,3,7] → [1,2,5,3,7] → [1,2,5,3] → [1,2,3]
  
  In the second example, f(A) = 2 and no additional operation is needed.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <class T>
  int getbit(T s, int i) {
    return (s >> i) & 1;
  }
  template <class T>
  T onbit(T s, int i) {
    return s | (T(1) << i);
  }
  template <class T>
  T offbit(T s, int i) {
    return s & (~(T(1) << i));
  }
  template <class T>
  int cntbit(T s) {
    return __builtin_popcount(s);
  }
  template <class T>
  T gcd(T a, T b) {
    T r;
    while (b != 0) {
      r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  template <class T>
  T lcm(T a, T b) {
    return a / gcd(a, b) * b;
  }
  int n, a[500005], f[500005];
  ;
  pair<int, int> P[500005];
  int get(int u) {
    int res = 0;
    for (int i = u; i > 0; i -= i & -i) {
      res = max(res, f[i]);
    }
    return res;
  }
  void update(int u, int gt) {
    for (int i = u; i < 500005; i += i & -i) f[i] = max(f[i], gt);
  }
  void solve() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n;
    for (int i = (1); i <= (n); ++i) {
      cin >> a[i];
      P[i] = make_pair(i + n - a[i], i);
    }
    sort(P + 1, P + n + 1);
    memset(f, 0, sizeof(f));
    int res = 0;
    update(1, 0);
    for (int i = (1); i <= (n); ++i) {
      int id = P[i].second;
      if (P[i].first < n) continue;
      int u = a[id];
      int A = get(u) + 1;
      res = max(res, A);
      update(u + 1, A);
    }
    cout << res << endl;
  }
  int main() {
    solve();
    return 0;
  }
  