#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
long long n, lis[200069];
pair<long long, long long> a[200069];
int main() {
  long long i, k, p, z = 0;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &k);
    k = i - k;
    a[i] = {i - k, -k};
    lis[i] = inf;
  }
  sort(a + 1, a + n + 1);
  for (i = 1; i <= n; i++) {
    k = -a[i].second;
    if (a[i].first > 0 && k >= 0) {
      p = upper_bound(lis + 1, lis + n + 1, k) - lis;
      z = max(z, p);
      lis[p] = k;
    }
  }
  printf("%lld\n", z);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long inf = 1e18;
long long n, lis[200069];
pair<long long, long long> a[200069];
int main() {
  long long i, k, p, z = 0;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &k);
    k = i - k;
    a[i] = {i - k, -k};
    lis[i] = inf;
  }
  sort(a + 1, a + n + 1);
  for (i = 1; i <= n; i++) {
    k = -a[i].second;
    if (a[i].first > 0 && k >= 0) {
      p = upper_bound(lis + 1, lis + n + 1, k) - lis;
      z = max(z, p);
      lis[p] = k;
    }
  }
  printf("%lld\n", z);
}

```