#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool upmax(T &x, T y) {
  return x < y ? (x = y, 1) : 0;
}
template <typename T>
inline bool upmin(T &x, T y) {
  return x > y ? (x = y, 1) : 0;
}
const int MAX_N = 2005;
int pre[MAX_N][2], suf[MAX_N][2];
int g[MAX_N][MAX_N][2][2];
int w[MAX_N], N;
int main() {
  scanf("%d", &N);
  for (int i = (0), _ = (N); i < _; ++i) {
    scanf("%d", &w[i]);
    w[i]--;
  }
  for (int i = (0), _ = (N); i < _; ++i) {
    if (i) memcpy(pre[i], pre[i - 1], sizeof pre[i]);
    for (int j = (2) - 1, _ = (w[i]); j >= _; --j)
      upmax(pre[i][j], pre[i][w[i]] + 1);
  }
  for (int i = (N)-1, _ = (0); i >= _; --i) {
    if (i < N - 1) memcpy(suf[i], suf[i + 1], sizeof suf[i]);
    for (int j = (0), _ = (w[i] + 1); j < _; ++j)
      upmax(suf[i][j], suf[i][w[i]] + 1);
  }
  int ans = pre[N - 1][1];
  for (int i = (0), _ = (N); i < _; ++i) {
    for (int a = (0), _ = (w[i] + 1); a < _; ++a)
      for (int b = (w[i]), _ = (2); b < _; ++b) {
        g[i][i][a][b] = 1;
      }
  }
  for (int l = (1), _ = (N); l < _; ++l)
    for (int i = (0), _ = (N - l); i < _; ++i) {
      int j = i + l;
      for (int l = (0), _ = (2); l < _; ++l)
        for (int a = (0), _ = (2 - l); a < _; ++a) {
          int b = a + l;
          g[i][j][a][b] =
              max((a < 1 ? g[i][j][a + 1][b] : 0), (b ? g[i][j][a][b - 1] : 0));
          upmax(g[i][j][a][b], g[i + 1][j][a][b] + (b == w[i]));
          upmax(g[i][j][a][b], g[i][j - 1][a][b] + (a == w[j]));
          upmax(ans, (i ? pre[i - 1][a] : 0) + g[i][j][a][b] +
                         (j < N - 1 ? suf[j + 1][b] : 0));
        }
    }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
template <typename T>
inline bool upmax(T &x, T y) {
  return x < y ? (x = y, 1) : 0;
}
template <typename T>
inline bool upmin(T &x, T y) {
  return x > y ? (x = y, 1) : 0;
}
const int MAX_N = 2005;
int pre[MAX_N][2], suf[MAX_N][2];
int g[MAX_N][MAX_N][2][2];
int w[MAX_N], N;
int main() {
  scanf("%d", &N);
  for (int i = (0), _ = (N); i < _; ++i) {
    scanf("%d", &w[i]);
    w[i]--;
  }
  for (int i = (0), _ = (N); i < _; ++i) {
    if (i) memcpy(pre[i], pre[i - 1], sizeof pre[i]);
    for (int j = (2) - 1, _ = (w[i]); j >= _; --j)
      upmax(pre[i][j], pre[i][w[i]] + 1);
  }
  for (int i = (N)-1, _ = (0); i >= _; --i) {
    if (i < N - 1) memcpy(suf[i], suf[i + 1], sizeof suf[i]);
    for (int j = (0), _ = (w[i] + 1); j < _; ++j)
      upmax(suf[i][j], suf[i][w[i]] + 1);
  }
  int ans = pre[N - 1][1];
  for (int i = (0), _ = (N); i < _; ++i) {
    for (int a = (0), _ = (w[i] + 1); a < _; ++a)
      for (int b = (w[i]), _ = (2); b < _; ++b) {
        g[i][i][a][b] = 1;
      }
  }
  for (int l = (1), _ = (N); l < _; ++l)
    for (int i = (0), _ = (N - l); i < _; ++i) {
      int j = i + l;
      for (int l = (0), _ = (2); l < _; ++l)
        for (int a = (0), _ = (2 - l); a < _; ++a) {
          int b = a + l;
          g[i][j][a][b] =
              max((a < 1 ? g[i][j][a + 1][b] : 0), (b ? g[i][j][a][b - 1] : 0));
          upmax(g[i][j][a][b], g[i + 1][j][a][b] + (b == w[i]));
          upmax(g[i][j][a][b], g[i][j - 1][a][b] + (a == w[j]));
          upmax(ans, (i ? pre[i - 1][a] : 0) + g[i][j][a][b] +
                         (j < N - 1 ? suf[j + 1][b] : 0));
        }
    }
  printf("%d\n", ans);
  return 0;
}

```