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 = 2 * 1e5 + 10, maxm = 2 * 8 * maxn, mod = 998244353;
  const int inf = 0x3f3f3f3f;
  const double pi = acos(-1), eps = 1e-7;
  int n;
  int a[maxn];
  int f[maxn];
  int tr[maxn];
  int lowbit(int x) { return x & (-x); }
  void modify(int x, int k) {
    while (x <= n) {
      tr[x] = max(tr[x], k);
      x += lowbit(x);
    }
  }
  int query(int x) {
    int res = 0;
    while (x) {
      res = max(res, tr[x]);
      x -= lowbit(x);
    }
    return res;
  }
  pair<int, int> b[maxn];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 1; i <= n; i++) {
      b[i] = {i - a[i], a[i]};
    }
    sort(b + 1, b + 1 + n);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (b[i].first >= 0) {
        f[i] = query(b[i].second) + 1;
        modify(b[i].second + 1, f[i]);
        ans = max(ans, f[i]);
      }
    }
    cout << ans << "\n";
    return 0;
  }
  