#include <bits/stdc++.h>
using namespace std;
const int NMax = 2e5 + 9;
pair<int, int> a[NMax];
int e[NMax];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) return (a.second > b.second);
  return (a.first < b.first);
}
void Solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i - a[i].first + 1;
  }
  sort(a, a + n, cmp);
  int m = 0;
  for (int i = 0; i < n; i++)
    if (a[i].second >= 0) {
      int f = upper_bound(e, e + m, a[i].second) - e;
      if (f == m) m++;
      e[f] = a[i].second;
    }
  cout << m;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solve();
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const int NMax = 2e5 + 9;
pair<int, int> a[NMax];
int e[NMax];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) return (a.second > b.second);
  return (a.first < b.first);
}
void Solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i - a[i].first + 1;
  }
  sort(a, a + n, cmp);
  int m = 0;
  for (int i = 0; i < n; i++)
    if (a[i].second >= 0) {
      int f = upper_bound(e, e + m, a[i].second) - e;
      if (f == m) m++;
      e[f] = a[i].second;
    }
  cout << m;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solve();
}

```