#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int n, ans;
int a[N], dp[N][2], pd[N][2], f[N][N][2][2];
int main() {
  int i, j, k, l;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), a[i]--;
  for (i = 1; i <= n; i++) {
    if (a[i] == 0) {
      dp[i][1] = dp[i - 1][1];
      dp[i][0] = dp[i - 1][0] + 1;
    } else {
      dp[i][1] = max(dp[i - 1][1], dp[i - 1][0]) + 1;
      dp[i][0] = dp[i - 1][0];
    }
  }
  for (i = n; i >= 1; i--) {
    if (a[i] == 0) {
      pd[i][1] = pd[i + 1][1];
      pd[i][0] = max(pd[i + 1][1], pd[i + 1][0]) + 1;
    } else {
      pd[i][1] = pd[i + 1][1] + 1;
      pd[i][0] = pd[i + 1][0];
    }
  }
  ans = max(dp[n][0], dp[n][1]);
  for (i = 1; i <= n; i++) {
    for (j = i; j <= n; j++) {
      for (k = 0; k < 2; k++)
        for (l = 0; l < 2; l++) f[i][j][k][l] = f[i][j - 1][k][l];
      f[i][j][a[j]][a[j]]++;
      if (a[j] == 0) {
        for (k = 0; k < 2; k++)
          f[i][j][1][0] = max(f[i][j][1][0], f[i][j - 1][1][k] + 1);
      }
    }
  }
  for (i = 1; i <= n; i++)
    for (j = i + 1; j <= n; j++)
      for (l = 0; l < 2; l++)
        for (k = l; k < 2; k++)
          ans = max(ans, dp[i - 1][l] + pd[j + 1][k] + f[i][j][k][l]);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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>
using namespace std;
const int N = 2005;
int n, ans;
int a[N], dp[N][2], pd[N][2], f[N][N][2][2];
int main() {
  int i, j, k, l;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), a[i]--;
  for (i = 1; i <= n; i++) {
    if (a[i] == 0) {
      dp[i][1] = dp[i - 1][1];
      dp[i][0] = dp[i - 1][0] + 1;
    } else {
      dp[i][1] = max(dp[i - 1][1], dp[i - 1][0]) + 1;
      dp[i][0] = dp[i - 1][0];
    }
  }
  for (i = n; i >= 1; i--) {
    if (a[i] == 0) {
      pd[i][1] = pd[i + 1][1];
      pd[i][0] = max(pd[i + 1][1], pd[i + 1][0]) + 1;
    } else {
      pd[i][1] = pd[i + 1][1] + 1;
      pd[i][0] = pd[i + 1][0];
    }
  }
  ans = max(dp[n][0], dp[n][1]);
  for (i = 1; i <= n; i++) {
    for (j = i; j <= n; j++) {
      for (k = 0; k < 2; k++)
        for (l = 0; l < 2; l++) f[i][j][k][l] = f[i][j - 1][k][l];
      f[i][j][a[j]][a[j]]++;
      if (a[j] == 0) {
        for (k = 0; k < 2; k++)
          f[i][j][1][0] = max(f[i][j][1][0], f[i][j - 1][1][k] + 1);
      }
    }
  }
  for (i = 1; i <= n; i++)
    for (j = i + 1; j <= n; j++)
      for (l = 0; l < 2; l++)
        for (k = l; k < 2; k++)
          ans = max(ans, dp[i - 1][l] + pd[j + 1][k] + f[i][j][k][l]);
  printf("%d\n", ans);
  return 0;
}

```