933_A. A Twisty Movement
A dragon symbolizes wisdom, power and wealth. On Lunar New Year's Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.

A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence a1, a2, ..., an.

Little Tommy is among them. He would like to choose an interval [l, r] (1 ≤ l ≤ r ≤ n), then reverse al, al + 1, ..., ar so that the length of the longest non-decreasing subsequence of the new sequence is maximum.

A non-decreasing subsequence is a sequence of indices p1, p2, ..., pk, such that p1 < p2 < ... < pk and ap1 ≤ ap2 ≤ ... ≤ apk. The length of the subsequence is k.

Input

The first line contains an integer n (1 ≤ n ≤ 2000), denoting the length of the original sequence.

The second line contains n space-separated integers, describing the original sequence a1, a2, ..., an (1 ≤ ai ≤ 2, i = 1, 2, ..., n).

Output

Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.

Examples

Input

4
1 2 1 2


Output

4


Input

10
1 1 2 2 2 1 1 2 2 1


Output

9

Note

In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.

In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.
```python
# -*- coding: utf-8 -*-
# Watashi wa ARUGONEKO wo tsukatteru. [algoneko.github.io]
# I hate python --RedCat
from math import *
import re

n = int(input())
a = list(map(int, input().split()))
d1, d2, d3, d4 = [0]*n, [0]*n, [0]*n, [0]*n
d1[0] = d3[0] = (1 if a[0] == 1 else 0)
d2[0] = d4[0] = (1 if a[0] == 2 else 0)
for i in range(1,n):
    d1[i] = d1[i-1]
    d2[i] = d2[i-1]
    d3[i] = d3[i-1]
    d4[i] = d4[i-1]
    if a[i] == 1:
        # * -> 1
        # 2 -> 3
        d1[i] =     d1[i - 1]             + 1
        d3[i] = max(d3[i - 1], d2[i - 1]) + 1
    if a[i] == 2:
        # 1 -> 2
        # 3 -> 4
        d2[i] = max(d2[i - 1], d1[i - 1]) + 1
        d4[i] = max(d4[i - 1], d3[i - 1]) + 1
print(max(max(d1), max(d2), max(d3), max(d4)))
```