#include <bits/stdc++.h>
using namespace std;
const int N = 2020;
inline char get(void) {
  static char buf[100000], *S = buf, *T = buf;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 100000, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
template <typename T>
inline void read(T &x) {
  static char c;
  x = 0;
  int sgn = 0;
  for (c = get(); c < '0' || c > '9'; c = get())
    if (c == '-') sgn = 1;
  for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
  if (sgn) x = -x;
}
int n, ans;
int a[N];
int pre[N], suf[N];
int dp[N][2];
int main(void) {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + (a[i] == 1);
  for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] + (a[i] == 2);
  for (int i = n; i >= 1; i--) {
    memset(dp, 0, sizeof dp);
    for (int j = i; j >= 1; j--) {
      if (a[j] == 1) {
        dp[j][0] = dp[j + 1][0] + 1;
        dp[j][1] = max(dp[j][0], dp[j + 1][1]);
      } else {
        dp[j][0] = dp[j + 1][0];
        dp[j][1] = max(dp[j + 1][0] + 1, dp[j + 1][1] + 1);
        dp[j][1] = max(dp[j][1], dp[j][0]);
      }
      ans = max(pre[j - 1] + dp[j][1] + suf[i + 1], ans);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 2020;
inline char get(void) {
  static char buf[100000], *S = buf, *T = buf;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 100000, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
template <typename T>
inline void read(T &x) {
  static char c;
  x = 0;
  int sgn = 0;
  for (c = get(); c < '0' || c > '9'; c = get())
    if (c == '-') sgn = 1;
  for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
  if (sgn) x = -x;
}
int n, ans;
int a[N];
int pre[N], suf[N];
int dp[N][2];
int main(void) {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + (a[i] == 1);
  for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] + (a[i] == 2);
  for (int i = n; i >= 1; i--) {
    memset(dp, 0, sizeof dp);
    for (int j = i; j >= 1; j--) {
      if (a[j] == 1) {
        dp[j][0] = dp[j + 1][0] + 1;
        dp[j][1] = max(dp[j][0], dp[j + 1][1]);
      } else {
        dp[j][0] = dp[j + 1][0];
        dp[j][1] = max(dp[j + 1][0] + 1, dp[j + 1][1] + 1);
        dp[j][1] = max(dp[j][1], dp[j][0]);
      }
      ans = max(pre[j - 1] + dp[j][1] + suf[i + 1], ans);
    }
  }
  cout << ans << endl;
  return 0;
}

```