#include<bits/stdc++.h>
using namespace std;
#define INF 2000000000
#define N_MAX 100010
int dp[N_MAX];
int main(void){
	
	int n;cin >> n;
	vector<int> a(n);
	for(int i=0;i < n;i++) cin >> a[i];
	
	fill(dp,dp+n,INF);

	for(int i = n-1;i >= 0;i--){
		*upper_bound(dp,dp+n,a[i]) = a[i];
	}
	cout << lower_bound(dp,dp+n,INF) - dp << endl;
	return 0;
}
### Prompt

Your challenge is to write 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 INF 2000000000
#define N_MAX 100010
int dp[N_MAX];
int main(void){
	
	int n;cin >> n;
	vector<int> a(n);
	for(int i=0;i < n;i++) cin >> a[i];
	
	fill(dp,dp+n,INF);

	for(int i = n-1;i >= 0;i--){
		*upper_bound(dp,dp+n,a[i]) = a[i];
	}
	cout << lower_bound(dp,dp+n,INF) - dp << endl;
	return 0;
}
```