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 long long sz = 2e5;
  long long a[sz + 5];
  int merge(int left, int right) { return max(left, right); }
  struct node1 {
    node1* left;
    node1* right;
    int val;
    node1() {
      val = 0;
      left = NULL;
      right = NULL;
    }
  };
  struct node2 {
    node1* root;
    node2* left;
    node2* right;
    node2() {
      root = new node1();
      left = NULL;
      right = NULL;
    }
    int get(node1*& v, int l, int r, int y1, int y2) {
      if (l == y1 && r == y2) {
        return v->val;
      }
      int m = (l + r) / 2;
      if (y2 <= m) {
        if (v->left == NULL) return 0;
        return get(v->left, l, m, y1, y2);
      } else if (y1 > m) {
        if (v->right == NULL) return 0;
        return get(v->right, m + 1, r, y1, y2);
      } else {
        return merge((v->left == NULL) ? 0 : get(v->left, l, m, y1, m),
                     (v->right == NULL) ? 0 : get(v->right, m + 1, r, m + 1, y2));
      }
    }
    void upd(node1*& v, int l, int r, int y, int val) {
      if (l == r && l == y) {
        v->val = max(v->val, val);
        return;
      }
      int m = (l + r) / 2;
      if (y <= m) {
        if (v->left == NULL) v->left = new node1();
        upd(v->left, l, m, y, val);
      } else {
        if (v->right == NULL) v->right = new node1();
        upd(v->right, m + 1, r, y, val);
      }
      v->val = merge((v->left == NULL) ? 0 : v->left->val,
                     (v->right == NULL) ? 0 : v->right->val);
    }
  };
  struct node3 {
    node2* root;
    node3() { root = new node2(); }
    int get(node2*& v, int l, int r, int x1, int x2, int y1, int y2) {
      if (l == x1 && r == x2) {
        return v->get(v->root, 0, sz, y1, y2);
      }
      int m = (l + r) / 2;
      if (x2 <= m) {
        if (v->left == NULL) v->left = new node2();
        return get(v->left, l, m, x1, x2, y1, y2);
      } else if (x1 > m) {
        if (v->right == NULL) v->right = new node2();
        return get(v->right, m + 1, r, x1, x2, y1, y2);
      } else {
        return merge(
            (v->left == NULL) ? 0 : get(v->left, l, m, x1, m, y1, y2),
            (v->right == NULL) ? 0 : get(v->right, m + 1, r, m + 1, x2, y1, y2));
      }
    }
    void upd(node2*& v, int l, int r, int x, int y, int val) {
      if (l == r && l == x) {
        v->upd(v->root, 0, sz, y, val);
        return;
      }
      int m = (l + r) / 2;
      if (x <= m) {
        if (v->left == NULL) v->left = new node2();
        upd(v->left, l, m, x, y, val);
      } else {
        if (v->right == NULL) v->right = new node2();
        upd(v->right, m + 1, r, x, y, val);
      }
      v->upd(v->root, 0, sz, y, val);
    }
  };
  node3* root;
  long long n;
  int main() {
    root = new node3();
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    root->upd(root->root, 0, sz, 0, 0, 0);
    for (int i = 1; i <= n; i++) {
      if (a[i] - i > 0) continue;
      long long new_ans =
          1 + root->get(root->root, 0, sz, 0, a[i] - 1, 0, i - a[i]);
      root->upd(root->root, 0, sz, a[i], i - a[i], new_ans);
    }
    cout << root->get(root->root, 0, sz, 0, sz, 0, sz) << "\n";
  }
  