#include <bits/stdc++.h>
using namespace std;
const int N = 200010, mo = 1e9 + 7;
int n;
pair<int, int> e[N];
int dp[N], cnt;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> e[i].first;
    e[i].second = -i;
  }
  sort(e + 1, e + 1 + n);
  for (int i = 1; i <= n; i++) {
    int d = -e[i].second - e[i].first;
    if (d < 0) continue;
    int pos = upper_bound(dp + 1, dp + 1 + cnt, d) - dp;
    if (pos > cnt) cnt++;
    dp[pos] = d;
  }
  cout << cnt << '\n';
  return 0;
}

### 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 int N = 200010, mo = 1e9 + 7;
int n;
pair<int, int> e[N];
int dp[N], cnt;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> e[i].first;
    e[i].second = -i;
  }
  sort(e + 1, e + 1 + n);
  for (int i = 1; i <= n; i++) {
    int d = -e[i].second - e[i].first;
    if (d < 0) continue;
    int pos = upper_bound(dp + 1, dp + 1 + cnt, d) - dp;
    if (pos > cnt) cnt++;
    dp[pos] = d;
  }
  cout << cnt << '\n';
  return 0;
}

```