#include <bits/stdc++.h>
using namespace std;
const int N = 2005, INF = 1e9;
int n, a[N], b[N], c[N], dp[N][N];
void precompute(int arr[], int val) {
  for (int i = 1; i <= n; i++) arr[i] = arr[i - 1] + (val == a[i]);
}
inline int sum(int arr[], int l, int r) {
  int ans = l > r ? 0 : arr[r] - arr[l - 1];
  assert(ans >= 0);
  return ans;
}
int go(int l, int r) {
  if (l > r) return 0;
  if (dp[l][r] != -INF) return dp[l][r];
  int ans = max(go(l, r - 1), go(l + 1, r));
  if (a[l] == 2) ans = max(ans, 1 + go(l + 1, r));
  if (a[r] == 1) ans = max(ans, 1 + go(l, r - 1));
  dp[l][r] = ans;
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  precompute(b, 1);
  precompute(c, 2);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j] = -INF;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) go(i, j);
  int ans = max(sum(b, 1, n), sum(c, 1, n));
  for (int i = 1; i < n; i++) ans = max(ans, sum(b, 1, i) + sum(c, i + 1, n));
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      int flip_ans = sum(b, 1, i - 1) + dp[i][j] + sum(c, j + 1, n);
      ans = max(ans, flip_ans);
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in CPP for 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, INF = 1e9;
int n, a[N], b[N], c[N], dp[N][N];
void precompute(int arr[], int val) {
  for (int i = 1; i <= n; i++) arr[i] = arr[i - 1] + (val == a[i]);
}
inline int sum(int arr[], int l, int r) {
  int ans = l > r ? 0 : arr[r] - arr[l - 1];
  assert(ans >= 0);
  return ans;
}
int go(int l, int r) {
  if (l > r) return 0;
  if (dp[l][r] != -INF) return dp[l][r];
  int ans = max(go(l, r - 1), go(l + 1, r));
  if (a[l] == 2) ans = max(ans, 1 + go(l + 1, r));
  if (a[r] == 1) ans = max(ans, 1 + go(l, r - 1));
  dp[l][r] = ans;
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  precompute(b, 1);
  precompute(c, 2);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j] = -INF;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) go(i, j);
  int ans = max(sum(b, 1, n), sum(c, 1, n));
  for (int i = 1; i < n; i++) ans = max(ans, sum(b, 1, i) + sum(c, i + 1, n));
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      int flip_ans = sum(b, 1, i - 1) + dp[i][j] + sum(c, j + 1, n);
      ans = max(ans, flip_ans);
    }
  }
  cout << ans;
  return 0;
}

```