#include <bits/stdc++.h>
using namespace std;
int dp[3][2002];
int one[2002];
int two[2002];
int pre1[2002];
int pre2[2002];
int mx[2002][2002];
int vs[2002];
int main() {
  int n;
  cin >> n;
  int a[n + 1];
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1)
      dp[1][i] = dp[1][i - 1] + 1;
    else
      dp[2][i] = 1 + max(dp[1][i - 1], dp[2][i - 1]);
    one[i] = one[i - 1];
    two[i] = two[i - 1];
    if (a[i] == 1)
      one[i] += 1;
    else
      two[i] += 1;
  }
  for (int i = n; i >= 1; --i) {
    pre1[i] = pre1[i + 1];
    pre2[i] = pre2[i + 1];
    if (a[i] == 1) pre1[i] += 1;
    if (a[i] == 2) pre2[i] += 1;
    vs[i] = pre1[i] - pre2[i];
  }
  for (int i = 1; i <= n; ++i) {
    mx[i][i] = vs[i];
    for (int j = i + 1; j <= n; ++j) mx[i][j] = max(mx[i][j - 1], vs[j]);
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) ans = max(ans, max(dp[1][i], dp[2][i]));
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      int curr = one[i - 1] + two[n] - two[j];
      int value = mx[i + 1][j] + pre2[i] - pre1[j + 1];
      ans = max(ans, curr + value);
    }
  }
  cout << ans;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int dp[3][2002];
int one[2002];
int two[2002];
int pre1[2002];
int pre2[2002];
int mx[2002][2002];
int vs[2002];
int main() {
  int n;
  cin >> n;
  int a[n + 1];
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1)
      dp[1][i] = dp[1][i - 1] + 1;
    else
      dp[2][i] = 1 + max(dp[1][i - 1], dp[2][i - 1]);
    one[i] = one[i - 1];
    two[i] = two[i - 1];
    if (a[i] == 1)
      one[i] += 1;
    else
      two[i] += 1;
  }
  for (int i = n; i >= 1; --i) {
    pre1[i] = pre1[i + 1];
    pre2[i] = pre2[i + 1];
    if (a[i] == 1) pre1[i] += 1;
    if (a[i] == 2) pre2[i] += 1;
    vs[i] = pre1[i] - pre2[i];
  }
  for (int i = 1; i <= n; ++i) {
    mx[i][i] = vs[i];
    for (int j = i + 1; j <= n; ++j) mx[i][j] = max(mx[i][j - 1], vs[j]);
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) ans = max(ans, max(dp[1][i], dp[2][i]));
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      int curr = one[i - 1] + two[n] - two[j];
      int value = mx[i + 1][j] + pre2[i] - pre1[j + 1];
      ans = max(ans, curr + value);
    }
  }
  cout << ans;
  return 0;
}

```