#include <bits/stdc++.h>
using namespace std;
const long long N = 2e3 + 10;
long long n, ans, ar[N], dp[2][N][N], psum[2][N];
long long solve(long long l, long long r) {
  long long res = max(psum[0][n] - psum[0][0], psum[1][n] - psum[1][0]);
  res = max(res, psum[0][r] - psum[0][0] + dp[0][r][n]);
  res =
      max(res, psum[0][l] - psum[0][0] + dp[1][l][r] + psum[1][n] - psum[1][r]);
  res = max(res, psum[1][l] - psum[1][0] + psum[1][n] - psum[1][l]);
  return res;
}
int32_t main() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> ar[i];
  for (long long i = 0; i < 2; i++)
    for (long long j = 1; j <= n; j++)
      psum[i][j] = psum[i][j - 1] + (ar[j - 1] == i + 1);
  for (long long i = 0; i < n; i++) dp[0][i][i + 1] = dp[1][i][i + 1] = 1;
  for (long long len = 2; len <= n; len++) {
    for (long long l = 0; l <= n - len; l++) {
      long long r = l + len;
      if (ar[r - 1] == 2)
        dp[0][l][r] = 1 + dp[0][l][r - 1];
      else
        dp[0][l][r] = max(psum[0][r] - psum[0][l], dp[0][l][r - 1]);
      if (ar[l] == 2)
        dp[1][l][r] = 1 + dp[1][l + 1][r];
      else
        dp[1][l][r] = max(psum[0][r] - psum[0][l], dp[1][l + 1][r]);
    }
  }
  for (long long l = 0; l < n; l++)
    for (long long r = l + 1; r <= n; r++) ans = max(ans, solve(l, r));
  cout << ans;
}

### Prompt

Your challenge is to write 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 long long N = 2e3 + 10;
long long n, ans, ar[N], dp[2][N][N], psum[2][N];
long long solve(long long l, long long r) {
  long long res = max(psum[0][n] - psum[0][0], psum[1][n] - psum[1][0]);
  res = max(res, psum[0][r] - psum[0][0] + dp[0][r][n]);
  res =
      max(res, psum[0][l] - psum[0][0] + dp[1][l][r] + psum[1][n] - psum[1][r]);
  res = max(res, psum[1][l] - psum[1][0] + psum[1][n] - psum[1][l]);
  return res;
}
int32_t main() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> ar[i];
  for (long long i = 0; i < 2; i++)
    for (long long j = 1; j <= n; j++)
      psum[i][j] = psum[i][j - 1] + (ar[j - 1] == i + 1);
  for (long long i = 0; i < n; i++) dp[0][i][i + 1] = dp[1][i][i + 1] = 1;
  for (long long len = 2; len <= n; len++) {
    for (long long l = 0; l <= n - len; l++) {
      long long r = l + len;
      if (ar[r - 1] == 2)
        dp[0][l][r] = 1 + dp[0][l][r - 1];
      else
        dp[0][l][r] = max(psum[0][r] - psum[0][l], dp[0][l][r - 1]);
      if (ar[l] == 2)
        dp[1][l][r] = 1 + dp[1][l + 1][r];
      else
        dp[1][l][r] = max(psum[0][r] - psum[0][l], dp[1][l + 1][r]);
    }
  }
  for (long long l = 0; l < n; l++)
    for (long long r = l + 1; r <= n; r++) ans = max(ans, solve(l, r));
  cout << ans;
}

```