#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
int spf[MAXN];
int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
void precal() {
  spf[1] = 1;
  for (int i = 2; i < MAXN; i++) spf[i] = i;
  for (int i = 4; i < MAXN; i += 2) spf[i] = 2;
  for (int i = 3; i * i < MAXN; i++) {
    if (spf[i] == i)
      for (int j = i * i; j < MAXN; j += i)
        if (spf[j] == j) spf[j] = i;
  }
}
vector<int> getFactorization(int x) {
  vector<int> ret;
  while (x != 1) {
    ret.push_back(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
long long exp(long long n, long long m) {
  long long a = 1;
  while (m) {
    if (m & 1) a = (a * n % 1000000007) % 1000000007;
    m /= 2;
    n = (n * n) % 1000000007;
  }
  return a;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, t = 1;
  while (t--) {
    cin >> n;
    long long a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    int dp[5];
    for (int i = 0; i < 5; i++) dp[i] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 4; j++)
        dp[j + 1] = max(dp[j], dp[j + 1] + (j % 2 + 1 == a[i]));
    }
    cout << dp[4] << "\n";
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 MAXN = 100005;
int spf[MAXN];
int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
void precal() {
  spf[1] = 1;
  for (int i = 2; i < MAXN; i++) spf[i] = i;
  for (int i = 4; i < MAXN; i += 2) spf[i] = 2;
  for (int i = 3; i * i < MAXN; i++) {
    if (spf[i] == i)
      for (int j = i * i; j < MAXN; j += i)
        if (spf[j] == j) spf[j] = i;
  }
}
vector<int> getFactorization(int x) {
  vector<int> ret;
  while (x != 1) {
    ret.push_back(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
long long exp(long long n, long long m) {
  long long a = 1;
  while (m) {
    if (m & 1) a = (a * n % 1000000007) % 1000000007;
    m /= 2;
    n = (n * n) % 1000000007;
  }
  return a;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, t = 1;
  while (t--) {
    cin >> n;
    long long a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    int dp[5];
    for (int i = 0; i < 5; i++) dp[i] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 4; j++)
        dp[j + 1] = max(dp[j], dp[j + 1] + (j % 2 + 1 == a[i]));
    }
    cout << dp[4] << "\n";
  }
  return 0;
}

```