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.
n = int(input())
*A, = map(int, input().split())

seq = []
cur = 1; cnt = 0
for a in A:
    if cur == a:
        cnt += 1
    else:
        seq.append(cnt)
        cur = a; cnt = 1
if cnt > 0:
    seq.append(cnt)
ans = max(seq)

l = len(seq)
dp = [[0]*4 for i in range(l+1)]
for i in range(l):
    one = i % 2 #12
    s = seq[i]
    for j in range(4):
        dp[i+1][j] = max(dp[i+1][j], dp[i][j])
        if not one: # 2
            if j % 2 == 0:
                dp[i+1][j] = max(dp[i+1][j], dp[i][j] + s)
            else:
                if j == 1:
                    dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j] + s)
        else: # 1
            if j % 2 == 1:
                dp[i+1][j] = max(dp[i+1][j], dp[i][j] + s)
            else:
                dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j] + s)
#    print(dp[i+1])
print(max(dp[l]))



