#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int n,a[N];
multiset<int> S;

int main()
{
	cin>>n;
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	S.clear();
	for (int i=1;i<=n;i++) {
		auto it=S.lower_bound(a[i]);  //查找第一个大于等于a[i]的值 
		if (it!=S.begin()) S.erase(--it);  //(--it)就是小于a[i]且最大的值，a[i]将会替换它 
		S.insert(a[i]);
	}
	cout<<S.size()<<endl;
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int N=1e5+10;
int n,a[N];
multiset<int> S;

int main()
{
	cin>>n;
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	S.clear();
	for (int i=1;i<=n;i++) {
		auto it=S.lower_bound(a[i]);  //查找第一个大于等于a[i]的值 
		if (it!=S.begin()) S.erase(--it);  //(--it)就是小于a[i]且最大的值，a[i]将会替换它 
		S.insert(a[i]);
	}
	cout<<S.size()<<endl;
	return 0;
}
```