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.
#t=int(input())
t=1
for _ in range(t):
    n=int(input())
    l=list(map(int,input().split()))
    dp=[[0 for j in range(4)] for i in range(n)]
    p1=[0]*(n+1)
    p2=[0]*(n+1)
    for i in range(n):
        p1[i+1]+=p1[i]
        p2[i+1]+=p2[i]
        if l[i]==1:
            p1[i+1]+=1
        else:
            p2[i+1]+=1
    if l[0]==1:        
        dp[0][0]=1
    else:
        dp[0][1]=1
    for i in range(1,n):
        for j in range(i-1,-1,-1):
            if l[i]==1:
                dp[i][0]=p1[i+1]
                dp[i][1]=max(dp[j][1],dp[i][1])
                if dp[j][1]!=0:
                    dp[i][2]=max(dp[j][1]+1,dp[j][2]+1,dp[i][2])
                else:
                    dp[i][2]=max(dp[j][2]+1,dp[i][2])
                dp[i][3]=max(dp[j][2],dp[i][3])
            else:
                dp[i][0]=max(dp[j][0],dp[i][0])
                if dp[j][0]!=0:
                    dp[i][1]=max(dp[j][1]+1,dp[j][0]+1,dp[i][1])
                else:
                    dp[i][1]=max(dp[j][1]+1,dp[i][1])
                dp[i][2]=dp[j][2]
                if dp[j][2]!=0:
                    dp[i][3]=max(dp[j][3]+1,dp[j][2]+1,dp[i][3])
                else:
                    dp[i][3]=max(dp[j][3]+1,dp[i][3])
    maxi=0
    for i in range(n):
        for j in range(4):
            maxi=max(maxi,dp[i][j])
    print(maxi)        