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 = 2007
n = int(input().strip())
a = list(map(int, input().strip().split()))
dp = [[0] * 5 for i in range(N)]
for i in range(n):
    if a[i] == 1:
        dp[i][1] = 1
    else:
        dp[i][2] = 1
for i in range(n):
    for j in range(i):
        if a[i] == 1:
            dp[i][1] = max(dp[i][1], dp[j][0] + 1)
            dp[i][1] = max(dp[i][1], dp[j][1] + 1)
            dp[i][2] = max(dp[i][2], dp[j][2])
            dp[i][3] = max(dp[i][3], dp[j][2] + 1)
            dp[i][3] = max(dp[i][3], dp[j][3] + 1)
            dp[i][4] = max(dp[i][4], dp[j][4])
        else:
            dp[i][1] = max(dp[i][1], dp[j][1])
            dp[i][2] = max(dp[i][2], dp[j][1] + 1)
            dp[i][2] = max(dp[i][2], dp[j][2] + 1)
            dp[i][3] = max(dp[i][3], dp[j][3])
            dp[i][4] = max(dp[i][4], dp[j][3] + 1)
            dp[i][4] = max(dp[i][4], dp[j][4] + 1)
res = 0
for i in range(n):
    for j in range(1, 5):
        res = max(res, dp[i][j])
print(res)