#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
void solve();
void read_input() {}
void solve() {}
int a[2005];
int left1[2005];
int right2[2005];
int dp[2005][2005][2];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    left1[i] = left1[i - 1];
    if (a[i - 1] == 1) {
      left1[i]++;
    }
  }
  for (int i = n; i >= 0; i--) {
    right2[i] = right2[i + 1];
    if (a[i - 1] == 2) {
      right2[i]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      int a1 = a[j - 1] == 1 ? 1 : 0;
      int a2 = a[j - 1] == 2 ? 1 : 0;
      dp[i][j][0] = max(dp[i][j - 1][0] + a1, dp[i][j - 1][1] + a2);
      dp[i][j][1] = dp[i][j - 1][1] + a2;
    }
  }
  int res = INT_MIN;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      res = max(res,
                left1[i - 1] + right2[j + 1] + max(dp[i][j][0], dp[i][j][1]));
    }
  }
  cout << res << '\n';
  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 long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
void solve();
void read_input() {}
void solve() {}
int a[2005];
int left1[2005];
int right2[2005];
int dp[2005][2005][2];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    left1[i] = left1[i - 1];
    if (a[i - 1] == 1) {
      left1[i]++;
    }
  }
  for (int i = n; i >= 0; i--) {
    right2[i] = right2[i + 1];
    if (a[i - 1] == 2) {
      right2[i]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      int a1 = a[j - 1] == 1 ? 1 : 0;
      int a2 = a[j - 1] == 2 ? 1 : 0;
      dp[i][j][0] = max(dp[i][j - 1][0] + a1, dp[i][j - 1][1] + a2);
      dp[i][j][1] = dp[i][j - 1][1] + a2;
    }
  }
  int res = INT_MIN;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      res = max(res,
                left1[i - 1] + right2[j + 1] + max(dp[i][j][0], dp[i][j][1]));
    }
  }
  cout << res << '\n';
  return 0;
}

```