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;
  const int MAXN = 2e5 + 10;
  struct BIT {
    int n;
    vector<int> bit;
    BIT(int _n) {
      n = _n;
      bit = vector<int>(n, -n);
    }
    int mx(int k) {
      int s = -n;
      k++;
      while (k >= 1) {
        s = max(s, bit[k]);
        k -= k & -k;
      }
      return s;
    }
    void update(int p, int value) {
      p++;
      while (p <= n) {
        bit[p] = max(bit[p], value);
        p += p & -p;
      }
    }
  };
  vector<int> v[MAXN];
  int main() {
    int n;
    cin >> n;
    BIT bit(MAXN);
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      v[x].push_back(i);
    }
    bit.update(0, 0);
    int ans = 0;
    for (int i = 1; i < MAXN; i++) {
      reverse(v[i].begin(), v[i].end());
      for (auto x : v[i]) {
        if (x < i) continue;
        int res = bit.mx(x - i) + 1;
        ans = max(ans, res);
        bit.update(x - i, res);
      }
    }
    cout << ans << "\n";
    return 0;
  }
  