#include<bits/stdc++.h>
using namespace std;
int n,B[100005],a[100005],len;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    B[1]=-a[1],len=1;
    for(int i=2;i<=n;i++){
      if(-a[i]>=B[len]) B[++len]=-a[i];
      else{
        int pos=upper_bound(B+1,B+len,-a[i])-B;
        B[pos]=-a[i];
      }
    }
    cout<<len<<endl;
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int n,B[100005],a[100005],len;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    B[1]=-a[1],len=1;
    for(int i=2;i<=n;i++){
      if(-a[i]>=B[len]) B[++len]=-a[i];
      else{
        int pos=upper_bound(B+1,B+len,-a[i])-B;
        B[pos]=-a[i];
      }
    }
    cout<<len<<endl;
    return 0;
}

```