#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e3 + 10;
int N;
int A[maxn];
int cnt[maxn][2];
int dp[maxn][maxn];
int ans[maxn];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= N; i++) {
    if (A[i] == 1) {
      cnt[i][0] = cnt[i - 1][0] + 1;
      cnt[i][1] = cnt[i - 1][1];
    } else {
      cnt[i][0] = cnt[i - 1][0];
      cnt[i][1] = cnt[i - 1][1] + 1;
    }
  }
  for (int i = 1; i <= N; i++) {
    ans[i] = 1;
    for (int j = 1; j < i; j++) {
      if (A[j] <= A[i]) {
        ans[i] = max(ans[i], ans[j] + 1);
      }
    }
  }
  memset(dp, 0, sizeof(dp));
  int res = ans[N];
  for (int i = N; i >= 1; i--) {
    for (int j = i - 1; j >= 1; j--) {
      if (A[j] == 1) {
        dp[j][i] = dp[j + 1][i];
      } else {
        int t = cnt[i][0] - cnt[j][0];
        dp[j][i] = max(t, dp[j + 1][i]) + 1;
      }
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = i + 1; j <= N; j++) {
      int t1 = cnt[i - 1][0];
      int t2 = cnt[N][1] - cnt[j][1];
      int tt = t1 + t2 + dp[i][j];
      if (tt > res) {
        res = tt;
      }
    }
  }
  printf("%d\n", res);
  return 0;
}

### Prompt

Generate 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 maxn = 2e3 + 10;
int N;
int A[maxn];
int cnt[maxn][2];
int dp[maxn][maxn];
int ans[maxn];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= N; i++) {
    if (A[i] == 1) {
      cnt[i][0] = cnt[i - 1][0] + 1;
      cnt[i][1] = cnt[i - 1][1];
    } else {
      cnt[i][0] = cnt[i - 1][0];
      cnt[i][1] = cnt[i - 1][1] + 1;
    }
  }
  for (int i = 1; i <= N; i++) {
    ans[i] = 1;
    for (int j = 1; j < i; j++) {
      if (A[j] <= A[i]) {
        ans[i] = max(ans[i], ans[j] + 1);
      }
    }
  }
  memset(dp, 0, sizeof(dp));
  int res = ans[N];
  for (int i = N; i >= 1; i--) {
    for (int j = i - 1; j >= 1; j--) {
      if (A[j] == 1) {
        dp[j][i] = dp[j + 1][i];
      } else {
        int t = cnt[i][0] - cnt[j][0];
        dp[j][i] = max(t, dp[j + 1][i]) + 1;
      }
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = i + 1; j <= N; j++) {
      int t1 = cnt[i - 1][0];
      int t2 = cnt[N][1] - cnt[j][1];
      int tt = t1 + t2 + dp[i][j];
      if (tt > res) {
        res = tt;
      }
    }
  }
  printf("%d\n", res);
  return 0;
}

```