#include <bits/stdc++.h>
using namespace std;
struct H {
  int i, x, y;
};
int k, m, n, b[200004];
H w[200004];
int C(H a, H b) { return a.x < b.x || a.x == b.x && a.i > b.i; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  fill(b + 1, b + n + 1, 1 << 30);
  for (int i = 1; i <= n; i++) {
    cin >> k;
    w[i] = {i, k, i - k};
  }
  sort(w + 1, w + n + 1, C);
  for (int i = 1; i <= n; i++) {
    if (w[i].y >= 0) b[upper_bound(b + 1, b + n + 1, w[i].y) - b] = w[i].y;
  }
  for (int i = 1; i <= n; i++) {
    if (b[i] < 1 << 30) m = i;
  }
  cout << m;
}

### Prompt

Your challenge is to write 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>
using namespace std;
struct H {
  int i, x, y;
};
int k, m, n, b[200004];
H w[200004];
int C(H a, H b) { return a.x < b.x || a.x == b.x && a.i > b.i; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  fill(b + 1, b + n + 1, 1 << 30);
  for (int i = 1; i <= n; i++) {
    cin >> k;
    w[i] = {i, k, i - k};
  }
  sort(w + 1, w + n + 1, C);
  for (int i = 1; i <= n; i++) {
    if (w[i].y >= 0) b[upper_bound(b + 1, b + n + 1, w[i].y) - b] = w[i].y;
  }
  for (int i = 1; i <= n; i++) {
    if (b[i] < 1 << 30) m = i;
  }
  cout << m;
}

```