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;
  template <typename T>
  T max3(T a, T b, T c) {
    return max(a, max(b, c));
  }
  template <typename T>
  T max4(T a, T b, T c, T d) {
    return max(max(a, d), max(b, c));
  }
  template <typename T>
  T min3(T a, T b, T c) {
    return min(a, min(b, c));
  }
  template <typename T>
  T min4(T a, T b, T c, T d) {
    return min(min(a, d), min(b, c));
  }
  template <typename T>
  T binex(T a, T b, T mod) {
    T ans = 1;
    a %= mod;
    while (b > 0) {
      if (b & 1) {
        ans *= a;
        ans %= mod;
      }
      a *= a;
      a %= mod;
      b >>= 1;
    }
    return ans;
  }
  template <class T>
  ostream &operator<<(ostream &os, vector<T> a) {
    for (auto x : a) {
      os << x << " ";
    }
    return os;
  }
  template <class T>
  ostream &operator<<(ostream &os, set<T> a) {
    for (auto x : a) {
      os << x << " ";
    }
    return os;
  }
  template <class T>
  ostream &operator<<(ostream &os, multiset<T> a) {
    for (auto x : a) {
      os << x << " ";
    }
    return os;
  }
  template <class T, class Q>
  ostream &operator<<(ostream &os, pair<T, Q> a) {
    os << "| ";
    os << a.first << ", " << a.second << " ";
    return os << "|";
  }
  template <class P, class Q, class T>
  ostream &operator<<(ostream &os, tuple<P, Q, T> a) {
    os << "| " << (get<0>(a)) << ", " << (get<1>(a)) << ", " << (get<2>(a))
       << "|";
    return os;
  }
  void precomp() {}
  bool comp(pair<int64_t, int64_t> &a, pair<int64_t, int64_t> &b) {
    if (a.first < b.first) {
      return true;
    } else if (a.first > b.first) {
      return false;
    } else {
      return a.second > b.second;
    }
  }
  void solve() {
    int64_t n;
    cin >> n;
    vector<pair<int64_t, int64_t>> a(n);
    for (int64_t i = 0; i < n; i++) {
      cin >> a[i].first;
      a[i].second = i + 1 - a[i].first;
    }
    sort(a.begin(), a.end(), comp);
    vector<int64_t> sto;
    sto.push_back(0);
    for (int64_t i = 0; i < n; i++) {
      if (a[i].second < 0) {
        continue;
      }
      if (sto.back() <= a[i].second) {
        sto.push_back(a[i].second);
      } else {
        int64_t ind = 0;
        while (ind < sto.size()) {
          if (sto[ind] <= a[i].second) {
            ind++;
          } else {
            break;
          }
        }
        sto[ind] = a[i].second;
      }
    }
    cout << sto.size() - 1 << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int64_t T;
    T = 1;
    precomp();
    for (int64_t I = 1; I <= T; I++) {
      solve();
    }
  }
  