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;
  using ll = long long;
  using ld = long double;
  using ull = unsigned long long;
  const int maxN = 1e6 + 1;
  const int mod = 1e9 + 7;
  const ll oo = 1e9;
  int n, a[maxN];
  pair<int, int> d[maxN];
  int f[maxN], e[maxN];
  inline void FastInput() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
  }
  void ReadInput() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      e[i] = oo;
      d[i] = {a[i], i - a[i]};
    }
    sort(d + 1, d + n + 1, [](pair<int, int> p, pair<int, int> q) {
      return (p.first < q.first || (p.first == q.first && p.second > q.second));
    });
  }
  void Solve() {
    for (int i = 1; i <= n; i++) {
      if (d[i].second < 0) continue;
      f[i] = upper_bound(e + 1, e + n + 1, d[i].second) - e;
      e[f[i]] = d[i].second;
    }
    cout << *max_element(f + 1, f + n + 1);
  }
  int main() {
    FastInput();
    ReadInput();
    Solve();
  }
  