#include <bits/stdc++.h>
using namespace std;
const int MAX = 2000 + 24;
int val[MAX];
int ed[MAX][3], st[MAX][3], cur[MAX][3][3];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &val[i]);
    }
    memset(ed, 0, sizeof(ed));
    for (int i = 1; i <= n; i++) {
      if (val[i] == 1) {
        ed[i][val[i]] = ed[i - 1][val[i]] + 1;
        ed[i][2] = ed[i - 1][2];
      } else {
        ed[i][val[i]] = max(ed[i - 1][val[i]] + 1, ed[i - 1][val[i] - 1] + 1);
        ed[i][1] = ed[i - 1][1];
      }
    }
    memset(st, 0, sizeof(st));
    for (int i = n; i > 0; i--) {
      if (val[i] == 1) {
        st[i][val[i]] = max(st[i + 1][val[i]] + 1, st[i + 1][2] + 1);
        st[i][2] = st[i + 1][2];
      } else {
        st[i][val[i]] = st[i + 1][val[i]] + 1;
        st[i][1] = st[i + 1][1];
      }
    }
    int result = 0;
    for (int i = 1; i <= n; i++) {
      memset(cur, 0, sizeof(cur));
      for (int j = i; j <= n; j++) {
        if (val[j] == 1) {
          cur[j][val[j]][1] = cur[j - 1][val[j]][1] + 1;
          cur[j][val[j]][2] =
              max(cur[j - 1][val[j]][2] + 1, cur[j - 1][2][2] + 1);
          cur[j][2][2] = cur[j - 1][2][2];
        } else {
          cur[j][val[j]][2] = cur[j - 1][val[j]][2] + 1;
          cur[j][1][1] = cur[j - 1][1][1];
          cur[j][1][2] = max(cur[j - 1][2][2], cur[j - 1][1][2]);
        }
        result = max(result, cur[j][1][1] + ed[i - 1][1] + st[j + 1][2]);
        result = max(result, cur[j][1][1] + ed[i - 1][1] + st[j + 1][1]);
        result = max(result, cur[j][1][2] + ed[i - 1][1] + st[j + 1][2]);
        result = max(result, cur[j][2][2] + ed[i - 1][2] + st[j + 1][2]);
      }
    }
    printf("%d\n", result);
  }
  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;
const int MAX = 2000 + 24;
int val[MAX];
int ed[MAX][3], st[MAX][3], cur[MAX][3][3];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &val[i]);
    }
    memset(ed, 0, sizeof(ed));
    for (int i = 1; i <= n; i++) {
      if (val[i] == 1) {
        ed[i][val[i]] = ed[i - 1][val[i]] + 1;
        ed[i][2] = ed[i - 1][2];
      } else {
        ed[i][val[i]] = max(ed[i - 1][val[i]] + 1, ed[i - 1][val[i] - 1] + 1);
        ed[i][1] = ed[i - 1][1];
      }
    }
    memset(st, 0, sizeof(st));
    for (int i = n; i > 0; i--) {
      if (val[i] == 1) {
        st[i][val[i]] = max(st[i + 1][val[i]] + 1, st[i + 1][2] + 1);
        st[i][2] = st[i + 1][2];
      } else {
        st[i][val[i]] = st[i + 1][val[i]] + 1;
        st[i][1] = st[i + 1][1];
      }
    }
    int result = 0;
    for (int i = 1; i <= n; i++) {
      memset(cur, 0, sizeof(cur));
      for (int j = i; j <= n; j++) {
        if (val[j] == 1) {
          cur[j][val[j]][1] = cur[j - 1][val[j]][1] + 1;
          cur[j][val[j]][2] =
              max(cur[j - 1][val[j]][2] + 1, cur[j - 1][2][2] + 1);
          cur[j][2][2] = cur[j - 1][2][2];
        } else {
          cur[j][val[j]][2] = cur[j - 1][val[j]][2] + 1;
          cur[j][1][1] = cur[j - 1][1][1];
          cur[j][1][2] = max(cur[j - 1][2][2], cur[j - 1][1][2]);
        }
        result = max(result, cur[j][1][1] + ed[i - 1][1] + st[j + 1][2]);
        result = max(result, cur[j][1][1] + ed[i - 1][1] + st[j + 1][1]);
        result = max(result, cur[j][1][2] + ed[i - 1][1] + st[j + 1][2]);
        result = max(result, cur[j][2][2] + ed[i - 1][2] + st[j + 1][2]);
      }
    }
    printf("%d\n", result);
  }
  return 0;
}

```