#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long best[2222][2222];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> a(n + 1), ones(n + 1), twos(n + 1);
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    ones[i] = ones[i - 1] + (a[i] == 1);
    twos[i] = twos[i - 1] + (a[i] == 2);
  }
  for (long long i = 1; i <= n; ++i) {
    best[i][i] = 1;
    long long t = (a[i] == 2);
    for (long long j = i - 1; j >= 1; --j) {
      best[i][j] = best[i][j + 1];
      if (a[j] == 2) {
        best[i][j] = max(best[i][j], t + 1);
        ++t;
      } else {
        ++best[i][j];
      }
    }
    t = (a[i] == 2);
    for (long long j = i + 1; j <= n; ++j) {
      best[i][j] = best[i][j - 1];
      if (a[j] == 2) {
        best[i][j] = max(best[i][j], t + 1);
        ++t;
      } else {
        ++best[i][j];
      }
    }
  }
  long long ans = 0;
  for (long long l = 1; l <= n; ++l)
    for (long long r = l; r <= n; ++r) {
      long long res1 = 0, res2 = 0, res3 = 0;
      if (l > 1) {
        res1 += best[l - 1][1];
        res2 += ones[l - 1];
        res3 += ones[l - 1];
      }
      res1 += twos[r] - twos[l - 1];
      res2 += best[l][r];
      res3 += ones[r] - ones[l - 1];
      if (r < n) {
        res1 += twos[n] - twos[r];
        res2 += twos[n] - twos[r];
        res3 += best[n][r + 1];
      }
      ans = max({ans, res1, res2, res3});
    }
  cout << ans << endl;
}

### 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long best[2222][2222];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> a(n + 1), ones(n + 1), twos(n + 1);
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    ones[i] = ones[i - 1] + (a[i] == 1);
    twos[i] = twos[i - 1] + (a[i] == 2);
  }
  for (long long i = 1; i <= n; ++i) {
    best[i][i] = 1;
    long long t = (a[i] == 2);
    for (long long j = i - 1; j >= 1; --j) {
      best[i][j] = best[i][j + 1];
      if (a[j] == 2) {
        best[i][j] = max(best[i][j], t + 1);
        ++t;
      } else {
        ++best[i][j];
      }
    }
    t = (a[i] == 2);
    for (long long j = i + 1; j <= n; ++j) {
      best[i][j] = best[i][j - 1];
      if (a[j] == 2) {
        best[i][j] = max(best[i][j], t + 1);
        ++t;
      } else {
        ++best[i][j];
      }
    }
  }
  long long ans = 0;
  for (long long l = 1; l <= n; ++l)
    for (long long r = l; r <= n; ++r) {
      long long res1 = 0, res2 = 0, res3 = 0;
      if (l > 1) {
        res1 += best[l - 1][1];
        res2 += ones[l - 1];
        res3 += ones[l - 1];
      }
      res1 += twos[r] - twos[l - 1];
      res2 += best[l][r];
      res3 += ones[r] - ones[l - 1];
      if (r < n) {
        res1 += twos[n] - twos[r];
        res2 += twos[n] - twos[r];
        res3 += best[n][r + 1];
      }
      ans = max({ans, res1, res2, res3});
    }
  cout << ans << endl;
}

```