#include <bits/stdc++.h>
using namespace std;
const int N = 200200;
const int mod = 1e9 + 7;
int n, a[N];
int pre[N], suf[N];
int dp[2030][2030];
inline int get1(int l, int r) {
  if (l > r) return 0;
  return r - l + 1 - (pre[r] - pre[l - 1]);
}
inline int get2(int l, int r) {
  if (l > r) return 0;
  return pre[r] - pre[l - 1];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) pre[i] = pre[i - 1] + (a[i] == 2);
  for (int i = n; i >= 1; --i) suf[i] = suf[i + 1] + (a[i] == 2);
  int ans = 1;
  for (int i = 1; i <= n + 1; ++i) ans = max(ans, get1(1, i - 1) + get2(i, n));
  for (int r = n; r >= 1; --r) {
    int ones = 0;
    for (int l = r; l >= 1; --l) {
      dp[l][r] = max(1, dp[l + 1][r] + (a[l] == 2));
      dp[l][r] = max(dp[l][r], ones + 1);
      if (a[l] == 1) ++ones;
    }
  }
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      int a = get1(1, l - 1);
      int b = get2(r + 1, n);
      ans = max(ans, a + b + dp[l][r]);
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Generate a cpp solution 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;
const int N = 200200;
const int mod = 1e9 + 7;
int n, a[N];
int pre[N], suf[N];
int dp[2030][2030];
inline int get1(int l, int r) {
  if (l > r) return 0;
  return r - l + 1 - (pre[r] - pre[l - 1]);
}
inline int get2(int l, int r) {
  if (l > r) return 0;
  return pre[r] - pre[l - 1];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) pre[i] = pre[i - 1] + (a[i] == 2);
  for (int i = n; i >= 1; --i) suf[i] = suf[i + 1] + (a[i] == 2);
  int ans = 1;
  for (int i = 1; i <= n + 1; ++i) ans = max(ans, get1(1, i - 1) + get2(i, n));
  for (int r = n; r >= 1; --r) {
    int ones = 0;
    for (int l = r; l >= 1; --l) {
      dp[l][r] = max(1, dp[l + 1][r] + (a[l] == 2));
      dp[l][r] = max(dp[l][r], ones + 1);
      if (a[l] == 1) ++ones;
    }
  }
  for (int l = 1; l <= n; ++l) {
    for (int r = l; r <= n; ++r) {
      int a = get1(1, l - 1);
      int b = get2(r + 1, n);
      ans = max(ans, a + b + dp[l][r]);
    }
  }
  cout << ans;
  return 0;
}

```