#include <bits/stdc++.h>
using namespace std;
int n,d[100005],l,a[100005];
int main(){
	scanf("%d",&n);
	for (int i=1; i<=n; i++){
		scanf("%d",&a[i]);
	}
	l=1;
	d[1]=a[n];
	for (int i=n-1; i>=1; i--){
		int tt=upper_bound(d+1,d+1+l,a[i])-d;
		if (tt>l){
			d[++l]=a[i];
		} 
		if (d[tt]>a[i]){
			d[tt]=a[i];
		} 
	}
	printf("%d",l);
}
### Prompt

Your challenge is to write 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>
using namespace std;
int n,d[100005],l,a[100005];
int main(){
	scanf("%d",&n);
	for (int i=1; i<=n; i++){
		scanf("%d",&a[i]);
	}
	l=1;
	d[1]=a[n];
	for (int i=n-1; i>=1; i--){
		int tt=upper_bound(d+1,d+1+l,a[i])-d;
		if (tt>l){
			d[++l]=a[i];
		} 
		if (d[tt]>a[i]){
			d[tt]=a[i];
		} 
	}
	printf("%d",l);
}
```