#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a;
  for (int i = 0; i < n; i++) {
    int val;
    cin >> val;
    int d = i + 1 - val;
    if (d >= 0) {
      a.push_back({d, val});
    }
  }
  sort(a.begin(), a.end());
  vector<int> b(a.size());
  for (int i = 0; i < a.size(); i++) {
    b[i] = a[i].second;
  }
  vector<int> tails;
  for (int i = 0; i < a.size(); i++) {
    int k = tails.size();
    int index =
        upper_bound(tails.begin(), tails.end(), b[i] - 1) - tails.begin() - 1;
    if (k == 0) {
      tails.push_back(b[i]);
    } else if (index == -1) {
      tails[0] = b[i];
    } else if (index == k - 1) {
      if (tails[k - 1] != b[i]) {
        tails.push_back(b[i]);
      }
    } else {
      tails[index + 1] = b[i];
    }
  }
  cout << tails.size() << endl;
}

### Prompt

Please create a solution in Cpp to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a;
  for (int i = 0; i < n; i++) {
    int val;
    cin >> val;
    int d = i + 1 - val;
    if (d >= 0) {
      a.push_back({d, val});
    }
  }
  sort(a.begin(), a.end());
  vector<int> b(a.size());
  for (int i = 0; i < a.size(); i++) {
    b[i] = a[i].second;
  }
  vector<int> tails;
  for (int i = 0; i < a.size(); i++) {
    int k = tails.size();
    int index =
        upper_bound(tails.begin(), tails.end(), b[i] - 1) - tails.begin() - 1;
    if (k == 0) {
      tails.push_back(b[i]);
    } else if (index == -1) {
      tails[0] = b[i];
    } else if (index == k - 1) {
      if (tails[k - 1] != b[i]) {
        tails.push_back(b[i]);
      }
    } else {
      tails[index + 1] = b[i];
    }
  }
  cout << tails.size() << endl;
}

```