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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimization("unroll-loops, no-stack-protector")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  mt19937_64 rnd;
  const long long maxn = 2e5 + 50;
  const long long mod = 1e9 + 7;
  const long long base = 1e13;
  const long long magic = 2000;
  set<pair<long long, long long> > st[4 * maxn];
  struct Seqment_tree {
    long long get(long long id, long long left, long long right, long long x,
                  long long y, long long h) {
      if (x > right || y < left) return -base;
      if (x <= left && y >= right) {
        auto it = st[id].lower_bound(make_pair(h, -1));
        long long ans = 0;
        if (it != st[id].begin()) {
          it--;
          ans = max(ans, (*it).second);
        }
        return ans;
      }
      long long mid = (left + right) / 2;
      return max(get(id * 2, left, mid, x, y, h),
                 get(id * 2 + 1, mid + 1, right, x, y, h));
    }
    void update(long long id, long long left, long long right, long long pos,
                long long h, long long diff) {
      if (pos < left || pos > right) return;
      auto it = st[id].lower_bound(make_pair(h, -1));
      bool chk = true;
      if (it != st[id].begin()) {
        it--;
        if ((*it).second >= diff) {
          chk = false;
        } else {
          it++;
        }
      }
      if (chk) {
        vector<pair<long long, long long> > vt;
        while (it != st[id].end()) {
          auto p = (*it);
          if (diff >= p.second) {
            vt.push_back(p);
            it++;
          } else
            break;
        }
        for (auto to : vt) st[id].erase(to);
        st[id].insert(make_pair(h, diff));
      }
      if (left == right) {
        return;
      }
      long long mid = (left + right) / 2;
      update(id * 2, left, mid, pos, h, diff);
      update(id * 2 + 1, mid + 1, right, pos, h, diff);
    }
  } man;
  long long dp[maxn];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    if (fopen("t.inp", "r")) {
      freopen("test.inp", "r", stdin);
      freopen("test.out", "w", stdout);
    }
    long long n;
    cin >> n;
    long long ans = 0;
    long long len = 2 * n;
    for (int i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      if (i >= x) {
        long long h = man.get(1, 1, n, 1, i - x + 1, x) + 1;
        ans = max(ans, h);
        man.update(1, 1, n, i - x + 1, x, h);
      }
    }
    cout << ans;
  }
  