#include <bits/stdc++.h>
#define sz(x) (int)x.size()
using namespace std;
typedef long long ll;

int main() {
	int n, a[100010];
	scanf("%d",&n);
	int c=0;
	multiset<int> st;
	for(int i=1; i<=n; i++) {
		scanf("%d",a+i);
		auto it=st.lower_bound(a[i]);
		if(it!=st.begin()) {
			it--;
			st.erase(st.find(*it));
		} else c++;
		st.insert(a[i]);
	}
	printf("%d",c);
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
#define sz(x) (int)x.size()
using namespace std;
typedef long long ll;

int main() {
	int n, a[100010];
	scanf("%d",&n);
	int c=0;
	multiset<int> st;
	for(int i=1; i<=n; i++) {
		scanf("%d",a+i);
		auto it=st.lower_bound(a[i]);
		if(it!=st.begin()) {
			it--;
			st.erase(st.find(*it));
		} else c++;
		st.insert(a[i]);
	}
	printf("%d",c);
	return 0;
}
```