#include<bits/stdc++.h>
#define maxn 200010

using namespace std;

int q[maxn],l,a[maxn],n;

int main() {
	int i,x;
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&a[i]);
	for (i=1;i<=n/2;i++) 
		swap(a[n-i+1],a[i]);
	for (i=1;i<=n;i++) {
		if (a[i]>=q[l]) q[++l]=a[i];
		else {
			x=upper_bound(q+1,q+l+1,a[i])-q;
			q[x]=a[i];
		}
	}
	printf("%d\n",l);
	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>
#define maxn 200010

using namespace std;

int q[maxn],l,a[maxn],n;

int main() {
	int i,x;
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&a[i]);
	for (i=1;i<=n/2;i++) 
		swap(a[n-i+1],a[i]);
	for (i=1;i<=n;i++) {
		if (a[i]>=q[l]) q[++l]=a[i];
		else {
			x=upper_bound(q+1,q+l+1,a[i])-q;
			q[x]=a[i];
		}
	}
	printf("%d\n",l);
	return 0;
}
```