#include <bits/stdc++.h>
using namespace std;
const int N = 2005, INF = 1e9;
int n, a[N], b[N][N], dp[N], c[N], d[N];
inline int sum(int prefix_sum[], int s, int e) {
  return prefix_sum[e] - prefix_sum[s - 1];
}
void precompute(int val, int arr[]) {
  for (int i = 1; i <= n; i++) arr[i] = arr[i - 1] + (a[i] == val);
}
int go(int x, int y) {
  if (x > y) return 0;
  if (b[x][y] != -INF) return b[x][y];
  b[x][y] = max(go(x, y - 1), go(x + 1, y));
  if (a[x] == 2) b[x][y] = max(b[x][y], 1 + go(x + 1, y));
  if (a[y] == 1) b[x][y] = max(b[x][y], 1 + go(x, y - 1));
  return b[x][y];
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) b[i][j] = -INF;
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      if (b[i][j] == -INF) go(i, j);
    }
  precompute(1, c);
  precompute(2, d);
  dp[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] == 2) {
      dp[i] = dp[i - 1] + 1;
    } else {
      dp[i] = max(dp[i - 1], sum(c, 1, i));
      for (int j = i - 1; j >= 1; j--) {
        dp[i] = max(dp[i], sum(c, 1, j - 1) + b[j][i]);
      }
    }
  }
  cout << dp[n];
  return 0;
}

### 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 int N = 2005, INF = 1e9;
int n, a[N], b[N][N], dp[N], c[N], d[N];
inline int sum(int prefix_sum[], int s, int e) {
  return prefix_sum[e] - prefix_sum[s - 1];
}
void precompute(int val, int arr[]) {
  for (int i = 1; i <= n; i++) arr[i] = arr[i - 1] + (a[i] == val);
}
int go(int x, int y) {
  if (x > y) return 0;
  if (b[x][y] != -INF) return b[x][y];
  b[x][y] = max(go(x, y - 1), go(x + 1, y));
  if (a[x] == 2) b[x][y] = max(b[x][y], 1 + go(x + 1, y));
  if (a[y] == 1) b[x][y] = max(b[x][y], 1 + go(x, y - 1));
  return b[x][y];
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) b[i][j] = -INF;
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      if (b[i][j] == -INF) go(i, j);
    }
  precompute(1, c);
  precompute(2, d);
  dp[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] == 2) {
      dp[i] = dp[i - 1] + 1;
    } else {
      dp[i] = max(dp[i - 1], sum(c, 1, i));
      for (int j = i - 1; j >= 1; j--) {
        dp[i] = max(dp[i], sum(c, 1, j - 1) + b[j][i]);
      }
    }
  }
  cout << dp[n];
  return 0;
}

```