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: Wrong Answer)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
using P   = pair<ll, ll>;
using mat = vector<vector<ll>>;
typedef string::const_iterator State;
class ParseError {};

void IOS() { ios::sync_with_stdio(false), cin.tie(0); }

const ll INF = 1e16;
const ll INM = 114514;
const ll MOD = 1e9 + 7;
const ld EPS = 1e-12;

template <typename T>
void dump(T x) { cout << x << endl; }
void dumpf(ld x, int t) { cout << setprecision(t) << fixed << x << endl; }
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
	fill((T *)array, (T *)(array + N), val);
}
inline ll mod(ll a, ll b) {
	if (a > 0) return a % b;
	if (a % b == 0) return 0;
	ll x = -a / b + 1;
	a += x * b;
	return a % b;
}
ll powm(ll a, ll b, ll c) {
	if (b == 0)
		return 1;
	else if (b % 2 == 0) {
		ll d = powm(a, b / 2, c);
		return (d * d) % c;
	} else
		return (a * powm(a, b - 1, c)) % c;
}
ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll dx[4] = {1, 0, -1, 0};
ll dy[4] = {0, 1, 0, -1};

int main() {
	IOS();
	ll n;
	cin >> n;
	vector<ll> a(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	ll ans = 0;
	set<ll, greater<ll>> s;
	for (int i = 0; i < n; i++) {
		auto it = s.upper_bound(a[i]);
		if (it == s.end()) {
			ans++;
		} else {
			s.erase(it);
		}
		s.insert(a[i]);
	}
	dump(ans);
}