#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 10;
int dp[N][N];
int a[N];
int c1[N], c2[N];
int solve(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (l == r)
    dp[l][r] = 1;
  else if (a[l] == 2)
    dp[l][r] = solve(l + 1, r) + 1;
  else
    dp[l][r] = max(solve(l + 1, r), c1[r] - c1[l] + 1);
  return dp[l][r];
}
int main() {
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      dp[i][j] = -1;
    }
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (a[0] == 1)
    c1[0] = 1, c2[0] = 0;
  else
    c1[0] = 0, c2[0] = 1;
  for (int i = 1; i < n; i++) {
    c1[i] = c1[i - 1];
    c2[i] = c2[i - 1];
    if (a[i] == 1)
      c1[i]++;
    else
      c2[i]++;
  }
  int ans = max(c2[n - 1], c1[n - 1]);
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (i == 0)
        ans = max(ans, solve(i, j) + c2[n - 1] - c2[j]);
      else
        ans = max(ans, c1[i - 1] + solve(i, j) + c2[n - 1] - c2[j]);
    }
  }
  cout << ans << endl;
  return 0;
}

### 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>
using namespace std;
const int N = 2e3 + 10;
int dp[N][N];
int a[N];
int c1[N], c2[N];
int solve(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (l == r)
    dp[l][r] = 1;
  else if (a[l] == 2)
    dp[l][r] = solve(l + 1, r) + 1;
  else
    dp[l][r] = max(solve(l + 1, r), c1[r] - c1[l] + 1);
  return dp[l][r];
}
int main() {
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      dp[i][j] = -1;
    }
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (a[0] == 1)
    c1[0] = 1, c2[0] = 0;
  else
    c1[0] = 0, c2[0] = 1;
  for (int i = 1; i < n; i++) {
    c1[i] = c1[i - 1];
    c2[i] = c2[i - 1];
    if (a[i] == 1)
      c1[i]++;
    else
      c2[i]++;
  }
  int ans = max(c2[n - 1], c1[n - 1]);
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (i == 0)
        ans = max(ans, solve(i, j) + c2[n - 1] - c2[j]);
      else
        ans = max(ans, c1[i - 1] + solve(i, j) + c2[n - 1] - c2[j]);
    }
  }
  cout << ans << endl;
  return 0;
}

```