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;
  class segment_tree {
    vector<long long> t;
    long size;
  
   public:
    segment_tree(long n) {
      if (1)
        t.assign(4 * n + 1, 0);
      else
        t.assign(4 * n + 1, (1LL << 60));
      size = n;
    }
    void initialize_array(vector<long long>& v) {
      initialize_with_array(1, 0, size - 1, v);
    }
    void initialize_with_array(long startpos, long l, long r,
                               vector<long long>& v) {
      if (l == r) {
        t[startpos] = v[l];
      } else {
        long m = (l + r) / 2;
        initialize_with_array(2 * startpos, l, m, v);
        initialize_with_array(2 * startpos + 1, m + 1, r, v);
        if (1 == 1)
          t[startpos] = max(t[startpos * 2], t[startpos * 2 + 1]);
        else
          t[startpos] = min(t[startpos * 2], t[startpos * 2 + 1]);
      }
    }
    void update(long index, long long val) {
      update_full(1, 0, size - 1, index, val);
    }
    void update_full(long startpos, long l, long r, long index, long long val) {
      if (l == r) {
        t[startpos] = val;
      } else {
        long m = (l + r) / 2;
        if (index <= m)
          update_full(2 * startpos, l, m, index, val);
        else
          update_full(2 * startpos + 1, m + 1, r, index, val);
        if (1)
          t[startpos] = max(t[startpos * 2], t[startpos * 2 + 1]);
        else
          t[startpos] = min(t[startpos * 2], t[startpos * 2 + 1]);
      }
    }
    long long query(long l, long r) {
      if (l > r) {
        if (1)
          return 0;
        else
          return (1LL << 60);
      }
      return query_full(1, 0, size - 1, l, r);
    }
    long long query_full(long startpos, long left, long right, long l, long r) {
      if ((left >= l) && (right <= r)) return t[startpos];
      long m = (left + right) / 2;
      long long ans;
      if (1)
        ans = -(1LL << 60);
      else
        ans = (1LL << 60);
      if (m >= l) {
        if (1)
          ans = max(ans, query_full(startpos * 2, left, m, l, r));
        else
          ans = min(ans, query_full(startpos * 2, left, m, l, r));
      }
      if (m + 1 <= r) {
        if (1)
          ans = max(ans, query_full(startpos * 2 + 1, m + 1, right, l, r));
        else
          ans = min(ans, query_full(startpos * 2 + 1, m + 1, right, l, r));
      }
      return ans;
    }
  };
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int MAXX = 200100, n, i, j, k, ans, x;
    cin >> n;
    vector<int> a(n + 1), dp(n + 1, 0);
    vector<pair<int, int>> plist;
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      if (i >= a[i]) plist.push_back({i - a[i], a[i]});
    }
    sort(plist.begin(), plist.end());
    segment_tree st(n + 1);
    ans = 0;
    for (auto p : plist) {
      x = st.query(0, p.second - 1);
      if (x + 1 > dp[p.second]) {
        dp[p.second] = x + 1;
        st.update(p.second, x + 1);
        ans = max(ans, x + 1);
      }
    }
    cout << ans << "\n";
    return 0;
  }
  