#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;scanf("%d",&n);
    vector<int> a(n),p(n,1e9+7);
    for(int i=0;i<n;i++)scanf("%d",&a[n-1-i]);
    for(int i=0;i<n;i++){
        *upper_bound(p.begin(),p.end(),a[i])=a[i];
    }
    int ans=lower_bound(p.begin(),p.end(),1e9+7)-p.begin();
    printf("%d\n",ans);
    return 0;
}
### Prompt

In cpp, your task is to solve 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;scanf("%d",&n);
    vector<int> a(n),p(n,1e9+7);
    for(int i=0;i<n;i++)scanf("%d",&a[n-1-i]);
    for(int i=0;i<n;i++){
        *upper_bound(p.begin(),p.end(),a[i])=a[i];
    }
    int ans=lower_bound(p.begin(),p.end(),1e9+7)-p.begin();
    printf("%d\n",ans);
    return 0;
}
```