#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin>>n;
    int a[n];
    for(int i=0;i<n;i++)cin>>a[i];
    vector<int>x(1,a[0]);
    for(int i=1;i<n;i++){
        if(x.back()>=a[i])x.push_back(a[i]);
        else x[upper_bound(x.begin(),x.end(),a[i],greater<int>())-x.begin()]=a[i];
    }
    cout<<x.size()<<endl;
    return 0;
}
### Prompt

Generate 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;
int main(){
    int n;
    cin>>n;
    int a[n];
    for(int i=0;i<n;i++)cin>>a[i];
    vector<int>x(1,a[0]);
    for(int i=1;i<n;i++){
        if(x.back()>=a[i])x.push_back(a[i]);
        else x[upper_bound(x.begin(),x.end(),a[i],greater<int>())-x.begin()]=a[i];
    }
    cout<<x.size()<<endl;
    return 0;
}
```