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.
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 9;
const long long mod = 1e9 + 7;
vector<bool> prime(MAX, 1);
vector<int> spf(MAX, 1), primes;
void sieve() {
  prime[0] = prime[1] = 0;
  spf[2] = 2;
  for (long long i = 4; i < MAX; i += 2) {
    spf[i] = 2;
    prime[i] = 0;
  }
  primes.push_back(2);
  for (long long i = 3; i < MAX; i += 2) {
    if (prime[i]) {
      primes.push_back(i);
      spf[i] = i;
      for (long long j = i * i; j < MAX; j += i) {
        prime[j] = 0;
        if (spf[j] == 1) {
          spf[j] = i;
        }
      }
    }
  }
}
template <typename T>
void in(T &x) {
  cin >> x;
}
template <typename T, typename U>
void in(T &x, U &y) {
  cin >> x >> y;
}
template <typename T, typename U, typename V>
void in(T &x, U &y, V &z) {
  cin >> x >> y >> z;
}
template <typename T>
void ps(T x) {
  cout << x << " ";
}
template <typename T>
void ps(const vector<T> &x, int n) {
  for (int i = 0; i < n; i++) {
    cout << x[i];
    (i == n - 1) ? cout << '\n' : cout << " ";
  }
}
template <typename T>
void pl(T x) {
  cout << x << '\n';
}
template <typename T>
void pl(const vector<T> &x, int n) {
  for (int i = 0; i < n; i++) {
    cout << x[i] << "\n";
  }
}
template <typename T>
T power(T a, T b) {
  T res = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b = b >> 1;
  }
  return res;
}
template <typename T>
T power(T a, T b, T m) {
  T res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % m;
    }
    a = (a * a) % m;
    b = b >> 1;
  }
  return res % m;
}
void virtual_main() {}
void real_main() {
  int n;
  cin >> n;
  vector<int> v(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  vector<vector<int> > dp(n + 1, vector<int>(5, 0));
  for (int i = 1; i <= n; i++) {
    if (v[i] == 1) {
      dp[i][1] = dp[i - 1][1] + 1;
      dp[i][3] = max({dp[i - 1][1], dp[i - 1][2], dp[i - 1][3]}) + 1;
      dp[i][2] = dp[i - 1][2];
      dp[i][4] = dp[i - 1][4];
    } else {
      dp[i][1] = dp[i - 1][1];
      dp[i][2] = max(dp[i - 1][1], dp[i - 1][2]) + 1;
      dp[i][3] = dp[i - 1][3];
      dp[i][4] =
          max({dp[i - 1][1], dp[i - 1][2], dp[i - 1][3], dp[i - 1][4]}) + 1;
    }
  }
  pl(max({dp[n][1], dp[n][2], dp[n][3], dp[n][4]}));
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  virtual_main();
  int test_cases = 1;
  for (int i = 1; i <= test_cases; i++) {
    real_main();
  }
  return 0;
}
