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>
  #pragma comment(linker, "/STACK:36777216")
  using namespace std;
  long long mod = 1e9 + 7;
  long double eps = 1e-12;
  long double pi = acosl(-1);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  ifstream in("input.txt");
  ofstream out("output.txt");
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  #pragma GCC optimization("unroll-loops")
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  using uc = unsigned char;
  using us = unsigned short;
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    vector<pair<int, int>> b;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] <= i + 1) b.push_back({i + 1 - a[i], a[i]});
      a[i] = i + 1 - a[i];
    }
    sort(b.begin(), b.end());
    if (b.size() == 0) {
      cout << 0;
      return;
    }
    vector<int> ans(1, b[0].second);
    for (int i = 1; i < b.size(); i++) {
      int id = lower_bound(ans.begin(), ans.end(), b[i].second) - ans.begin();
      if (id == ans.size()) {
        ans.push_back(b[i].second);
      } else {
        ans[id] = b[i].second;
      }
    }
    cout << ans.size();
  }
  int main() {
    cout << setprecision(20) << fixed;
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    clock_t t1 = clock();
    int t = 1;
    int cs = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  