#include<bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
using namespace std;

int main(){
	int n;
	cin >> n;
	vector<int> a(n);
	rep(i, n) cin >> a[i];
	deque<int> d;
	rep(i,n) {
		int p = lower_bound(d.begin(), d.end(), a[i]) - d.begin();
		if(p==0){
			d.push_front(a[i]);
		}
		else d[p-1] = a[i];
	}
	int ans = d.size();
	cout << ans << endl;
	return 0;
}
### Prompt

Generate a cpp solution to the following problem:

You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:

* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.



Find the minimum number of colors required to satisfy the condition.

Constraints

* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N
A_1
:
A_N


Output

Print the minimum number of colors required to satisfy the condition.

Examples

Input

5
2
1
4
5
3


Output

2


Input

4
0
0
0
0


Output

4

### Response

```cpp
#include<bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
using namespace std;

int main(){
	int n;
	cin >> n;
	vector<int> a(n);
	rep(i, n) cin >> a[i];
	deque<int> d;
	rep(i,n) {
		int p = lower_bound(d.begin(), d.end(), a[i]) - d.begin();
		if(p==0){
			d.push_front(a[i]);
		}
		else d[p-1] = a[i];
	}
	int ans = d.size();
	cout << ans << endl;
	return 0;
}
```