#include <bits/stdc++.h>
long long int suf[2001][2];
long long int pre[2001][2];
long long int sm[2001][2001];
long long int dp[2001][2001];
long long int a[10000];
long long int max(long long int a, long long int b);
long long int func(int index, int l, int flag);
int main() {
  int n;
  scanf("%d", &n);
  int i, j;
  for (i = 1; i <= n; ++i) {
    sm[i][i] = -LLONG_MAX;
    scanf("%lld", &a[i]);
    a[i] = a[i] - 1;
  }
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) sm[i][j] = -1;
  if (a[1] == 1)
    pre[1][1] = 1;
  else
    pre[1][0] = 1;
  for (i = 2; i <= n; ++i) {
    if (a[i] == 1) {
      pre[i][1] = pre[i - 1][1] + 1;
      pre[i][0] = pre[i - 1][0];
    } else {
      pre[i][1] = pre[i - 1][1];
      pre[i][0] = pre[i - 1][0] + 1;
    }
  }
  if (a[n] == 1)
    suf[n][1] = 1;
  else
    suf[n][0] = 1;
  for (i = n - 1; i >= 1; --i) {
    if (a[i] == 1) {
      suf[i][1] = suf[i + 1][1] + 1;
      suf[i][0] = suf[i + 1][0];
    } else {
      suf[i][1] = suf[i + 1][1];
      suf[i][0] = suf[i + 1][0] + 1;
    }
  }
  for (i = 1; i <= n; ++i) {
    for (j = i + 1; j <= n; ++j) {
      if (a[j] == 0) {
        sm[i][j] = max(1 + sm[i][j - 1], 1 + suf[i][1] - suf[j][1]);
      } else {
        sm[i][j] = sm[i][j - 1];
      }
    }
  }
  long long int ans = -LLONG_MAX;
  for (i = 1; i <= n; ++i) {
    for (j = i; j <= n; ++j) {
      ans = max(ans, pre[i - 1][0] + sm[i][j] + suf[j + 1][1]);
      ans = max(ans, pre[i][0] + suf[i + 1][1]);
      ans = max(ans, pre[i - 1][0] + suf[i][1]);
    }
  }
  printf("%lld\n", ans);
  return 0;
}
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
long long int suf[2001][2];
long long int pre[2001][2];
long long int sm[2001][2001];
long long int dp[2001][2001];
long long int a[10000];
long long int max(long long int a, long long int b);
long long int func(int index, int l, int flag);
int main() {
  int n;
  scanf("%d", &n);
  int i, j;
  for (i = 1; i <= n; ++i) {
    sm[i][i] = -LLONG_MAX;
    scanf("%lld", &a[i]);
    a[i] = a[i] - 1;
  }
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) sm[i][j] = -1;
  if (a[1] == 1)
    pre[1][1] = 1;
  else
    pre[1][0] = 1;
  for (i = 2; i <= n; ++i) {
    if (a[i] == 1) {
      pre[i][1] = pre[i - 1][1] + 1;
      pre[i][0] = pre[i - 1][0];
    } else {
      pre[i][1] = pre[i - 1][1];
      pre[i][0] = pre[i - 1][0] + 1;
    }
  }
  if (a[n] == 1)
    suf[n][1] = 1;
  else
    suf[n][0] = 1;
  for (i = n - 1; i >= 1; --i) {
    if (a[i] == 1) {
      suf[i][1] = suf[i + 1][1] + 1;
      suf[i][0] = suf[i + 1][0];
    } else {
      suf[i][1] = suf[i + 1][1];
      suf[i][0] = suf[i + 1][0] + 1;
    }
  }
  for (i = 1; i <= n; ++i) {
    for (j = i + 1; j <= n; ++j) {
      if (a[j] == 0) {
        sm[i][j] = max(1 + sm[i][j - 1], 1 + suf[i][1] - suf[j][1]);
      } else {
        sm[i][j] = sm[i][j - 1];
      }
    }
  }
  long long int ans = -LLONG_MAX;
  for (i = 1; i <= n; ++i) {
    for (j = i; j <= n; ++j) {
      ans = max(ans, pre[i - 1][0] + sm[i][j] + suf[j + 1][1]);
      ans = max(ans, pre[i][0] + suf[i + 1][1]);
      ans = max(ans, pre[i - 1][0] + suf[i][1]);
    }
  }
  printf("%lld\n", ans);
  return 0;
}
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}

```