#include<bits/stdc++.h>
using namespace std;
#define reg register
#define M 200005
int fir,sec;
int A[M],n;
int B[M],m;
int main() {
	scanf("%d",&n);
	for(int i=1; i<=n; i++)scanf("%d",A+i);
	for(int i=n; i>=1; --i) {
		int t=upper_bound(B+1,B+1+m,A[i])-B;
		if(t==m+1)B[++m]=A[i];
		else B[t]=A[i];
	}
	printf("%d",m);
}


### Prompt

Your task is to create 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;
#define reg register
#define M 200005
int fir,sec;
int A[M],n;
int B[M],m;
int main() {
	scanf("%d",&n);
	for(int i=1; i<=n; i++)scanf("%d",A+i);
	for(int i=n; i>=1; --i) {
		int t=upper_bound(B+1,B+1+m,A[i])-B;
		if(t==m+1)B[++m]=A[i];
		else B[t]=A[i];
	}
	printf("%d",m);
}


```