#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using ii = pair<int, int>;
using vii = vector<ii>;
using vvii = vector<vii>;
const int MOD = 1000000007;
const int MX = 100001;
int main() {
  int n;
  cin >> n;
  vii v;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (i >= x) v.push_back(ii(i + 1 - x, x));
  }
  sort(v.begin(), v.end());
  vi s;
  for (auto i : v) {
    auto it = lower_bound(s.begin(), s.end(), i.second);
    if (it == s.end())
      s.push_back(i.second);
    else
      *it = i.second;
  }
  cout << (int)s.size();
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
using vi = vector<int>;
using vvi = vector<vi>;
using ii = pair<int, int>;
using vii = vector<ii>;
using vvii = vector<vii>;
const int MOD = 1000000007;
const int MX = 100001;
int main() {
  int n;
  cin >> n;
  vii v;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (i >= x) v.push_back(ii(i + 1 - x, x));
  }
  sort(v.begin(), v.end());
  vi s;
  for (auto i : v) {
    auto it = lower_bound(s.begin(), s.end(), i.second);
    if (it == s.end())
      s.push_back(i.second);
    else
      *it = i.second;
  }
  cout << (int)s.size();
  return 0;
}

```