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 ULL = unsigned long long;
  using VI = vector<int>;
  using VL = vector<LL>;
  using PII = pair<int, int>;
  using PLL = pair<LL, LL>;
  void dout() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dout(Head H, Tail... T) {
    cerr << H << ' ';
    dout(T...);
  }
  const int INF = 1e9 + 47;
  struct BIT {
    BIT(int n) : bit(n + 1, -INF) {}
    void setmax(int pos, int val) {
      for (int i = pos + 1; i < (int)bit.size(); i += i & -i)
        bit[i] = max(bit[i], val);
    }
    int getmax(int pos) {
      int res = -INF;
      for (int i = pos + 1; i > 0; i -= i & -i) res = max(res, bit[i]);
      return res;
    }
    vector<int> bit;
  };
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    VI a(n);
    vector<PII> b;
    for (int i = (0); i < (n); ++i) {
      cin >> a[i];
      b.emplace_back(i + 1 - a[i], a[i]);
    }
    sort(b.begin(), b.end());
    BIT bit(n + 1);
    bit.setmax(0, 0);
    int ans = 0;
    for (auto [dif, x] : b) {
      if (dif < 0) continue;
      int dp = bit.getmax(x - 1) + 1;
      bit.setmax(x, dp);
      ans = max(ans, dp);
    }
    cout << ans << "\n";
    return 0;
  }
  