#include <bits/stdc++.h>
using namespace std;
int n, a[2007], b[2007], beg;
int lisu[2007][2007], lisd[2007][2007];
int lis1(int start) {
  int s[2], l[2007];
  memset(s, 0, sizeof(s));
  memset(l, 0, sizeof(l));
  int ret = 0;
  s[a[start]]++;
  l[start]++;
  for (int i = start + 1; i < n; i++) {
    int r = s[0];
    if (a[i]) r = max(r, s[1]);
    l[i] = max(1, r + 1);
    s[a[i]] = max(s[a[i]], l[i]);
    ret = max(ret, l[i]);
    lisu[start][i] = ret;
  }
  return ret;
}
int lis2(int start) {
  int s[2], l[2007];
  memset(s, 0, sizeof(s));
  memset(l, 0, sizeof(l));
  int ret = 0;
  s[b[start]]++;
  l[start]++;
  for (int i = start + 1; i < n; i++) {
    int r = s[0];
    if (b[i]) r = max(r, s[1]);
    l[i] = max(1, r + 1);
    s[b[i]] = max(s[b[i]], l[i]);
    ret = max(ret, l[i]);
    lisd[n - 1 - i][n - 1 - start] = ret;
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], --a[i], b[i] = a[i];
  reverse(b, b + n);
  for (int i = 0; i < n; i++) lis1(i), lis2(i);
  int rk = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) rk = max(rk, lisd[i][j] - lisu[i][j]);
  std::cout << max(1, rk + lisu[0][n - 1]) << '\n';
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 n, a[2007], b[2007], beg;
int lisu[2007][2007], lisd[2007][2007];
int lis1(int start) {
  int s[2], l[2007];
  memset(s, 0, sizeof(s));
  memset(l, 0, sizeof(l));
  int ret = 0;
  s[a[start]]++;
  l[start]++;
  for (int i = start + 1; i < n; i++) {
    int r = s[0];
    if (a[i]) r = max(r, s[1]);
    l[i] = max(1, r + 1);
    s[a[i]] = max(s[a[i]], l[i]);
    ret = max(ret, l[i]);
    lisu[start][i] = ret;
  }
  return ret;
}
int lis2(int start) {
  int s[2], l[2007];
  memset(s, 0, sizeof(s));
  memset(l, 0, sizeof(l));
  int ret = 0;
  s[b[start]]++;
  l[start]++;
  for (int i = start + 1; i < n; i++) {
    int r = s[0];
    if (b[i]) r = max(r, s[1]);
    l[i] = max(1, r + 1);
    s[b[i]] = max(s[b[i]], l[i]);
    ret = max(ret, l[i]);
    lisd[n - 1 - i][n - 1 - start] = ret;
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], --a[i], b[i] = a[i];
  reverse(b, b + n);
  for (int i = 0; i < n; i++) lis1(i), lis2(i);
  int rk = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) rk = max(rk, lisd[i][j] - lisu[i][j]);
  std::cout << max(1, rk + lisu[0][n - 1]) << '\n';
}

```