#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 2100;
int a[N];
int dp1[N][N][2][2];
int dp2[N][N][2][2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    a[i]--;
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      for (int st = 0; st <= 1; st++)
        for (int ed = st; ed <= 1; ed++)
          dp1[i][j][st][ed] = dp1[i][j - 1][st][ed];
      for (int st = 0; st <= 1; st++)
        for (int ed = st; ed <= a[j]; ed++)
          dp1[i][j][st][a[j]] =
              max(dp1[i][j][st][a[j]], dp1[i][j - 1][st][ed] + 1);
    }
  for (int j = n; j >= 1; j--)
    for (int i = j; i >= 1; i--) {
      for (int st = 0; st <= 1; st++)
        for (int ed = st; ed <= 1; ed++)
          dp2[i][j][st][ed] = dp2[i + 1][j][st][ed];
      for (int st = 0; st <= 1; st++)
        for (int ed = st; ed <= a[i]; ed++)
          dp2[i][j][st][a[i]] =
              max(dp2[i][j][st][a[i]], dp2[i + 1][j][st][ed] + 1);
    }
  int ans = 0;
  for (int st = 0; st <= 1; st++)
    for (int ed = st; ed <= 1; ed++) ans = max(ans, dp1[1][n][st][ed]);
  for (int l = 1; l <= n; l++)
    for (int r = l; r <= n; r++) {
      for (int a = 0; a <= 1; a++)
        for (int b = a; b <= 1; b++)
          for (int c = b; c <= 1; c++)
            for (int d = c; d <= 1; d++)
              for (int e = d; e <= 1; e++)
                for (int f = e; f <= 1; f++)
                  ans = max(ans, dp1[1][l - 1][a][b] + dp2[l][r][c][d] +
                                     dp1[r + 1][n][e][f]);
    }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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 inf = 0x3f3f3f3f;
const int N = 2100;
int a[N];
int dp1[N][N][2][2];
int dp2[N][N][2][2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    a[i]--;
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      for (int st = 0; st <= 1; st++)
        for (int ed = st; ed <= 1; ed++)
          dp1[i][j][st][ed] = dp1[i][j - 1][st][ed];
      for (int st = 0; st <= 1; st++)
        for (int ed = st; ed <= a[j]; ed++)
          dp1[i][j][st][a[j]] =
              max(dp1[i][j][st][a[j]], dp1[i][j - 1][st][ed] + 1);
    }
  for (int j = n; j >= 1; j--)
    for (int i = j; i >= 1; i--) {
      for (int st = 0; st <= 1; st++)
        for (int ed = st; ed <= 1; ed++)
          dp2[i][j][st][ed] = dp2[i + 1][j][st][ed];
      for (int st = 0; st <= 1; st++)
        for (int ed = st; ed <= a[i]; ed++)
          dp2[i][j][st][a[i]] =
              max(dp2[i][j][st][a[i]], dp2[i + 1][j][st][ed] + 1);
    }
  int ans = 0;
  for (int st = 0; st <= 1; st++)
    for (int ed = st; ed <= 1; ed++) ans = max(ans, dp1[1][n][st][ed]);
  for (int l = 1; l <= n; l++)
    for (int r = l; r <= n; r++) {
      for (int a = 0; a <= 1; a++)
        for (int b = a; b <= 1; b++)
          for (int c = b; c <= 1; c++)
            for (int d = c; d <= 1; d++)
              for (int e = d; e <= 1; e++)
                for (int f = e; f <= 1; f++)
                  ans = max(ans, dp1[1][l - 1][a][b] + dp2[l][r][c][d] +
                                     dp1[r + 1][n][e][f]);
    }
  printf("%d\n", ans);
  return 0;
}

```