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)

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//------------------------------------------------------------------------------
//------------------------------------------------------------------------------
#define index_of(as, x) \
  distance(as.begin(), lower_bound(as.begin(), as.end(), x))
vector<int> lis_fast(const vector<int>& a) {
  const int n = a.size();
  vector<int> A(n, inf);
  vector<int> id(n);
  for (int i = 0; i < n; ++i) {
    id[i] = index_of(A, a[i]);
    A[ id[i] ] = a[i];
  }
  int m = *max_element(id.begin(), id.end());
  vector<int> b(m+1);
  for (int i = n-1; i >= 0; --i)
    if (id[i] == m) b[m--] = a[i];
  return b;
}

void _main()
{
    int N;
    cin >> N;
    vector<int> A(N);
    rep(i, 0, N) cin >> A[i];

    int sum = 0;
    int ans = 0;
    while (sum < N)
    {
        const vector<int> lis = lis_fast(A);
        sum += lis.size();
        int i = 0;
        vector<int> B;
        rep(j, 0, A.size())
        {
            if ((i < lis.size()) && (A[j] == lis[i]))
            {
                A[j] = inf;
                ++i;
            }
        }
        ++ans;
    }
    cout << ans << endl;
}