#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  vector<pair<int, int>> vec;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    int orig = v[i];
    v[i] = (i + 1) - v[i];
    if (v[i] >= 0) {
      vec.push_back(make_pair(v[i], orig));
    }
  }
  sort(vec.begin(), vec.end());
  vector<int> pos;
  for (pair<int, int> i : vec) {
    pos.push_back(i.second);
  }
  vector<int> ans;
  n = pos.size();
  for (int i = 0; i < n; i++) {
    auto it = lower_bound(ans.begin(), ans.end(), pos[i]);
    if (it == ans.end()) {
      ans.push_back(pos[i]);
    } else {
      *it = pos[i];
    }
  }
  cout << ans.size() << endl;
}

### Prompt

Generate a Cpp solution 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>
#pragma GCC optimize("O3")
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  vector<pair<int, int>> vec;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    int orig = v[i];
    v[i] = (i + 1) - v[i];
    if (v[i] >= 0) {
      vec.push_back(make_pair(v[i], orig));
    }
  }
  sort(vec.begin(), vec.end());
  vector<int> pos;
  for (pair<int, int> i : vec) {
    pos.push_back(i.second);
  }
  vector<int> ans;
  n = pos.size();
  for (int i = 0; i < n; i++) {
    auto it = lower_bound(ans.begin(), ans.end(), pos[i]);
    if (it == ans.end()) {
      ans.push_back(pos[i]);
    } else {
      *it = pos[i];
    }
  }
  cout << ans.size() << endl;
}

```