Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

5
2
1
4
5
3

Sample Output 1

2

We can satisfy the condition with two colors by, for example, painting 2 and 3 red and painting 1, 4, and 5 blue.

Sample Input 2

4
0
0
0
0

Sample Output 2

4

We have to paint all the integers with distinct colors.

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
#define loop(n, i) for(int i=0;i<n;i++)
#define loop1(n, i) for(int i=1;i<=n;i++)
#define rloop(n, i) for(int i=n;i;i++)
#define HERE cout << "HERE: " << __LINE__ << endl;
#define INSP(v) cout << v << " at " << __LINE__ << endl;
using namespace std;
using ll = long long;
template <class T>
using V = vector<T>;

int n;
V<int> A;

int main() {
    cin >> n;
    A.resize(n);
    loop (n, i) cin >> A[i];

    V<int> B(1, -1);
    loop (n, i) {
        auto itr = upper_bound(B.begin(), B.end(), A[i], greater<int>());
        if (itr == B.end()) {
            B.push_back(A[i]);
            sort(B.begin(), B.end(), greater<int>());
        } else {
            *itr = A[i];
        }

        // for (int k : B) {
        //     cout << k << " ";
        // }
        // cout << endl;
    }

    cout << B.size() << endl;
    return 0;
}