#include <bits/stdc++.h>
using namespace std;
inline long long Max(long long a, long long b) { return a > b ? a : b; }
inline long long Min(long long a, long long b) { return a < b ? a : b; }
inline long long read() {
  long long x = 0;
  int ch = getchar(), f = 1;
  while (!isdigit(ch) && (ch != '-') && (ch != EOF)) ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  return x * f;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10)
    write(x / 10), putchar(x % 10 + '0');
  else
    putchar(x + '0');
}
int n, a[2005], dp[5][2005], b[5] = {0, 1, 2, 1, 2};
int main() {
  n = read();
  for (int i = (int)(1); i <= (int)(n); i++) a[i] = read();
  for (int i = (int)(1); i <= (int)(4); i++)
    for (int j = (int)(1); j <= (int)(n); j++)
      dp[i][j] = Max(dp[i][j - 1] + (b[i] == a[j]), dp[i - 1][j]);
  write(dp[4][n]);
  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;
inline long long Max(long long a, long long b) { return a > b ? a : b; }
inline long long Min(long long a, long long b) { return a < b ? a : b; }
inline long long read() {
  long long x = 0;
  int ch = getchar(), f = 1;
  while (!isdigit(ch) && (ch != '-') && (ch != EOF)) ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  return x * f;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10)
    write(x / 10), putchar(x % 10 + '0');
  else
    putchar(x + '0');
}
int n, a[2005], dp[5][2005], b[5] = {0, 1, 2, 1, 2};
int main() {
  n = read();
  for (int i = (int)(1); i <= (int)(n); i++) a[i] = read();
  for (int i = (int)(1); i <= (int)(4); i++)
    for (int j = (int)(1); j <= (int)(n); j++)
      dp[i][j] = Max(dp[i][j - 1] + (b[i] == a[j]), dp[i - 1][j]);
  write(dp[4][n]);
  return 0;
}

```