341_B. Bubble Sort Graph
Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a1, a2, ..., an in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode). 
    
    
      
    procedure bubbleSortGraph()  
        build a graph G with n vertices and 0 edges  
        repeat  
            swapped = false  
            for i = 1 to n - 1 inclusive do:  
                if a[i] > a[i + 1] then  
                    add an undirected edge in G between a[i] and a[i + 1]  
                    swap( a[i], a[i + 1] )  
                    swapped = true  
                end if  
            end for  
        until not swapped   
        /* repeat the algorithm as long as swapped value is true. */   
    end procedure  
    

For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 105). The next line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n).

Output

Output a single integer — the answer to the problem. 

Examples

Input

3
3 1 2


Output

2

Note

Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2].
```python
from bisect import bisect_left, bisect_right, insort
R = lambda: map(int, input().split())
n, arr = int(input()), list(R())
dp = []
for i in range(n):
    idx = bisect_left(dp, arr[i])
    if idx >= len(dp):
        dp.append(arr[i])
    else:
        dp[idx] = arr[i]
print(len(dp))
```