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 + 10;
  struct item {
    int mx;
  };
  struct segtree {
    int sz;
    vector<item> values;
    void init(int n) {
      sz = 1;
      while (sz < n) sz *= 2;
      values.resize(2 * sz, NEUTRAL_ELEMENT);
    }
    item NEUTRAL_ELEMENT = {0};
    item merge(item a, item b) { return {max(a.mx, b.mx)}; }
    item single(int v) { return {v}; }
    void build(vector<int> &a, int x, int lx, int rx) {
      if (rx - lx == 1) {
        if (lx < (int)a.size()) {
          values[x] = single(a[lx]);
        }
        return;
      }
      int m = (lx + rx) / 2;
      build(a, 2 * x + 1, lx, m);
      build(a, 2 * x + 2, m, rx);
      values[x] = merge(values[2 * x + 1], values[2 * x + 2]);
    }
    void build(vector<int> &a) { build(a, 0, 0, sz); }
    void set(int i, int v, int x, int lx, int rx) {
      if (rx - lx == 1) {
        values[x] = merge(single(v), values[x]);
        return;
      }
      int m = (lx + rx) / 2;
      if (i < m) {
        set(i, v, 2 * x + 1, lx, m);
      } else {
        set(i, v, 2 * x + 2, m, rx);
      }
      values[x] = merge(values[2 * x + 1], values[2 * x + 2]);
    }
    void set(int i, int v) { set(i, v, 0, 0, sz); }
    item calc(int l, int r, int x, int lx, int rx) {
      if (l >= rx || lx >= r) return NEUTRAL_ELEMENT;
      if (lx >= l && rx <= r) return values[x];
      int m = (lx + rx) / 2;
      item s1 = calc(l, r, 2 * x + 1, lx, m);
      item s2 = calc(l, r, 2 * x + 2, m, rx);
      return merge(s1, s2);
    }
    item calc(int l, int r) { return calc(l, r, 0, 0, sz); }
  };
  void solve() {
    int n;
    cin >> n;
    int a[n + 1];
    a[0] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    vector<int> v[n + 1];
    for (int i = n; i >= 1; i--) {
      if (i - a[i] + 1 >= 0) v[i - a[i] + 1].push_back(i);
    }
    segtree st;
    st.init(N);
    for (int i = n; i >= 1; i--) {
      sort(v[i].begin(), v[i].end());
      reverse(v[i].begin(), v[i].end());
      for (int j : v[i]) {
        int mx = st.calc(a[j] + 1, N).mx;
        st.set(a[j], mx + 1);
      }
    }
    cout << st.calc(0, N).mx << "\n";
  }
  int main() {
    cin.tie(0);
    cin.sync_with_stdio(0);
    int t;
    t = 1;
    while (t--) {
      solve();
    }
  }
  