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 = 2e5 + 5;
  struct info {
    int maxi = 0;
  };
  struct segtree {
    int n;
    vector<info> t;
    segtree(int n) {
      this->n = n + 1;
      t.resize(n * 4);
    }
    void Merge(info &node, info &l, info &r) { node.maxi = max(l.maxi, r.maxi); }
    void update(int node, int l, int r, int i, int j, int add) {
      if (l > j || r < i) {
        return;
      }
      if (l >= i && r <= j) {
        t[node].maxi = add;
        return;
      }
      int mid = (l + r) / 2;
      update(node * 2, l, mid, i, j, add);
      update(node * 2 + 1, mid + 1, r, i, j, add);
      Merge(t[node], t[node * 2], t[node * 2 + 1]);
    }
    void update(int l, int add) { update(1, 0, n, l, l, add); }
    info query(int node, int l, int r, int i, int j) {
      if (l > j || r < i) {
        info x;
        return x;
      }
      if (l >= i && r <= j) {
        return t[node];
      }
      int mid = (l + r) / 2;
      info x = query(node * 2, l, mid, i, j);
      info y = query(node * 2 + 1, mid + 1, r, i, j);
      info ret;
      Merge(ret, x, y);
      return ret;
    }
    int query(int l, int r) {
      info ret = query(1, 0, n, l, r);
      return ret.maxi;
    }
  };
  int n;
  vector<pair<pair<int, int>, int> > a;
  segtree seg(N);
  int dp[N];
  bool comp(pair<pair<int, int>, int> &a, pair<pair<int, int>, int> &b) {
    if (a.first.first == b.first.first) {
      return a.first.second > b.first.second;
    }
    return a.first.first < b.first.first;
  }
  void solve(int l, int r) {
    if (l > r) {
      return;
    }
    if (l == r) {
      dp[l] = max(dp[l], 1);
      return;
    }
    int m = (l + r) / 2;
    solve(l, m);
    vector<pair<pair<int, int>, int> > now;
    for (int i = l; i <= r; i++) {
      now.push_back(a[i]);
    }
    sort(now.begin(), now.end(), comp);
    for (int i = 0; i < now.size(); i++) {
      int x = now[i].first.first;
      int y = now[i].first.second;
      int id = now[i].second;
      if (id >= l && id <= m) {
        seg.update(y, dp[id]);
      } else {
        int q = seg.query(0, y);
        dp[id] = max(dp[id], q + 1);
      }
    }
    for (int i = 0; i < now.size(); i++) {
      int x = now[i].first.first;
      int y = now[i].first.second;
      int id = now[i].second;
      if (id >= l && id <= m) {
        seg.update(y, 0);
      }
    }
    solve(m + 1, r);
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    for (int i = 1; i <= n; i++) {
      int x, y;
      cin >> x;
      y = i - x;
      if (y >= 0) {
        int sz = a.size();
        a.push_back({{x, y}, sz});
      }
    }
    n = a.size();
    if (n == 0) {
      cout << "0\n";
      return 0;
    };
    solve(0, n - 1);
    int ans = 0;
    for (int i = 0; i < n; i++) {
      ans = max(ans, dp[i]);
    }
    cout << ans << "\n";
    return 0;
  }
  