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 N = 4e5 + 10;
  namespace Seg {
  int f[N << 1];
  inline void Init() {
    for (int i = 0; i < N << 1; ++i) f[i] = -1e9;
  }
  inline void Update(int x, int y) {
    for (int i = N + x; i; i >>= 1) f[i] = max(f[i], y);
  }
  inline int Ask(int x, int y) {
    int ans = -1e9;
    for (int l = N + x - 1, r = N + y + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
      if (~l & 1) ans = max(ans, f[l ^ 1]);
      if (r & 1) ans = max(ans, f[r ^ 1]);
    }
    return ans;
  }
  }  // namespace Seg
  int n;
  pair<int, int> p[N];
  int main() {
    scanf("%d", &n);
    for (int i = 1, x; i <= n; ++i) {
      scanf("%d", &x);
      p[i] = pair<int, int>(i - x, x);
    }
    sort(p + 1, p + n + 1);
    Seg::Init();
    Seg::Update(0, 0);
    int ans = 0;
    for (int i = 1; i <= n; ++i)
      if (p[i].first >= 0) {
        int f = Seg::Ask(0, p[i].second - 1) + 1;
        ans = max(ans, f);
        Seg::Update(p[i].second, f);
      }
    printf("%d\n", ans);
    return 0;
  }
  