#include<cstdio>
#include<algorithm>
using namespace std;

#define ran 101101

int n;
int a[ran], f[ran];

int main() {
	scanf("%d", &n);
	for(int i=0;i<n;i++) scanf("%d", &a[i]);
	
	int h = 0;
	for(int i=n-1;i>=0;i--) {
		int p = upper_bound(f, f+h, a[i]) - f;
		if (p>=h) f[h++] = a[i];
		else f[p] = a[i];
	}
	printf("%d\n", h);
	
	return 0;
}
### Prompt

Develop a solution in Cpp 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<cstdio>
#include<algorithm>
using namespace std;

#define ran 101101

int n;
int a[ran], f[ran];

int main() {
	scanf("%d", &n);
	for(int i=0;i<n;i++) scanf("%d", &a[i]);
	
	int h = 0;
	for(int i=n-1;i>=0;i--) {
		int p = upper_bound(f, f+h, a[i]) - f;
		if (p>=h) f[h++] = a[i];
		else f[p] = a[i];
	}
	printf("%d\n", h);
	
	return 0;
}
```