#include <bits/stdc++.h>
using namespace std;
inline int rd() {
  char ch = getchar();
  int i = 0, f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    i = (i << 1) + (i << 3) + ch - '0';
    ch = getchar();
  }
  return i * f;
}
inline void W(int x) {
  static int buf[50];
  if (!x) {
    putchar('-');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  while (x) {
    buf[++buf[0]] = x % 10;
    x /= 10;
  }
  while (buf[0]) {
    putchar(buf[buf[0]--] + '0');
  }
}
const int N = 2e3 + 50;
int n, a[N], s[3][N];
int main() {
  n = rd();
  for (int i = 1; i <= n; i++)
    a[i] = rd(), s[1][i] = s[1][i - 1] + (a[i] == 1),
    s[2][i] = s[2][i - 1] + a[i] - 1;
  int ans = 0;
  for (int p = 0; p <= n; p++) {
    int mx1 = 0, mx2 = 0;
    for (int j = p; j <= n; j++)
      mx2 = max(mx2, s[1][j] - s[1][p] + s[2][n] - s[2][j]);
    for (int j = p; j >= 0; j--) mx1 = max(mx1, s[2][p] - s[2][j] + s[1][j]);
    ans = max(ans, mx1 + mx2);
  }
  cout << ans << endl;
}

### Prompt

Your task is to create 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 int rd() {
  char ch = getchar();
  int i = 0, f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    i = (i << 1) + (i << 3) + ch - '0';
    ch = getchar();
  }
  return i * f;
}
inline void W(int x) {
  static int buf[50];
  if (!x) {
    putchar('-');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  while (x) {
    buf[++buf[0]] = x % 10;
    x /= 10;
  }
  while (buf[0]) {
    putchar(buf[buf[0]--] + '0');
  }
}
const int N = 2e3 + 50;
int n, a[N], s[3][N];
int main() {
  n = rd();
  for (int i = 1; i <= n; i++)
    a[i] = rd(), s[1][i] = s[1][i - 1] + (a[i] == 1),
    s[2][i] = s[2][i - 1] + a[i] - 1;
  int ans = 0;
  for (int p = 0; p <= n; p++) {
    int mx1 = 0, mx2 = 0;
    for (int j = p; j <= n; j++)
      mx2 = max(mx2, s[1][j] - s[1][p] + s[2][n] - s[2][j]);
    for (int j = p; j >= 0; j--) mx1 = max(mx1, s[2][p] - s[2][j] + s[1][j]);
    ans = max(ans, mx1 + mx2);
  }
  cout << ans << endl;
}

```