#include<bits/stdc++.h>
using namespace std;
#define M 100005
int n,A[M],B[M],ans;
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>A[i];
	B[++ans]=A[n];
	for(int i=n-1;i>=1;i--){
		if(B[ans]<=A[i])B[++ans]=A[i];
		else {
			int l=1,r=ans,res;
			while(l<=r){
				int mid=l+r>>1;
				if(B[mid]<=A[i])l=mid+1;
				else r=mid-1,res=mid;
			}
			B[res]=A[i];
		}
	}
	cout<<ans;
	return 0;
}
### Prompt

Please create a solution in cpp 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;
#define M 100005
int n,A[M],B[M],ans;
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>A[i];
	B[++ans]=A[n];
	for(int i=n-1;i>=1;i--){
		if(B[ans]<=A[i])B[++ans]=A[i];
		else {
			int l=1,r=ans,res;
			while(l<=r){
				int mid=l+r>>1;
				if(B[mid]<=A[i])l=mid+1;
				else r=mid-1,res=mid;
			}
			B[res]=A[i];
		}
	}
	cout<<ans;
	return 0;
}
```