#include <bits/stdc++.h>
using namespace std;
vector<int> adj[2];
int dp[2003][6], sz = 0, tz = 0;
vector<int> v;
int solve(int idx, int sta) {
  if (dp[idx][sta] != -1) return dp[idx][sta];
  int e, f, g, h, p, q, r;
  vector<int>::iterator it;
  if (sta == 1 || sta == 3) {
    p = 1;
    if (adj[0][sz - 1] > idx) {
      it = upper_bound(adj[0].begin(), adj[0].end(), idx);
      h = it - adj[0].begin();
      r = adj[0][h];
      p = max(p, 1 + solve(r, sta));
    }
    if (tz > 0) {
      if (adj[1][tz - 1] > idx) {
        it = upper_bound(adj[1].begin(), adj[1].end(), idx);
        h = it - adj[1].begin();
        r = adj[1][h];
        p = max(p, 1 + solve(r, sta + 1));
      }
    }
    dp[idx][sta] = p;
    return dp[idx][sta];
  }
  if (sta == 2 || sta == 4) {
    p = 1;
    if (adj[1][tz - 1] > idx) {
      it = upper_bound(adj[1].begin(), adj[1].end(), idx);
      h = it - adj[1].begin();
      r = adj[1][h];
      p = max(p, 1 + solve(r, sta));
    }
    if (sz > 0 && sta != 4) {
      if (adj[0][sz - 1] > idx) {
        it = upper_bound(adj[0].begin(), adj[0].end(), idx);
        h = it - adj[0].begin();
        r = adj[0][h];
        p = max(p, 1 + solve(r, sta + 1));
      }
    }
    dp[idx][sta] = p;
    return dp[idx][sta];
  }
}
int main() {
  int n, i, j, k, a, b, c, ans = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a);
    v.push_back(a);
    adj[a - 1].push_back(i);
    if (a == 1)
      sz++;
    else {
      tz++;
    }
  }
  memset(dp, -1, sizeof dp);
  for (i = 0; i < n; i++) {
    ans = max(ans, solve(i, v[i]));
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
vector<int> adj[2];
int dp[2003][6], sz = 0, tz = 0;
vector<int> v;
int solve(int idx, int sta) {
  if (dp[idx][sta] != -1) return dp[idx][sta];
  int e, f, g, h, p, q, r;
  vector<int>::iterator it;
  if (sta == 1 || sta == 3) {
    p = 1;
    if (adj[0][sz - 1] > idx) {
      it = upper_bound(adj[0].begin(), adj[0].end(), idx);
      h = it - adj[0].begin();
      r = adj[0][h];
      p = max(p, 1 + solve(r, sta));
    }
    if (tz > 0) {
      if (adj[1][tz - 1] > idx) {
        it = upper_bound(adj[1].begin(), adj[1].end(), idx);
        h = it - adj[1].begin();
        r = adj[1][h];
        p = max(p, 1 + solve(r, sta + 1));
      }
    }
    dp[idx][sta] = p;
    return dp[idx][sta];
  }
  if (sta == 2 || sta == 4) {
    p = 1;
    if (adj[1][tz - 1] > idx) {
      it = upper_bound(adj[1].begin(), adj[1].end(), idx);
      h = it - adj[1].begin();
      r = adj[1][h];
      p = max(p, 1 + solve(r, sta));
    }
    if (sz > 0 && sta != 4) {
      if (adj[0][sz - 1] > idx) {
        it = upper_bound(adj[0].begin(), adj[0].end(), idx);
        h = it - adj[0].begin();
        r = adj[0][h];
        p = max(p, 1 + solve(r, sta + 1));
      }
    }
    dp[idx][sta] = p;
    return dp[idx][sta];
  }
}
int main() {
  int n, i, j, k, a, b, c, ans = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a);
    v.push_back(a);
    adj[a - 1].push_back(i);
    if (a == 1)
      sz++;
    else {
      tz++;
    }
  }
  memset(dp, -1, sizeof dp);
  for (i = 0; i < n; i++) {
    ans = max(ans, solve(i, v[i]));
  }
  printf("%d\n", ans);
  return 0;
}

```