#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e3 + 5;
const int INF = -0x3f3f3f3f;
int n, ai[maxn], f[maxn][maxn], g[maxn][maxn][2], dp[maxn][2][2];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ai[i]);
    ai[i]--;
  }
  memset(g, -0x3f, sizeof(g));
  memset(f, -0x3f, sizeof(f));
  for (int i = 1; i <= n; i++) {
    if (ai[i] == 1)
      g[i][i][1] = 1;
    else
      f[i][i] = 1;
    for (int v = i + 1; v <= n; v++) {
      if (ai[v] == 0) f[i][v] = 1;
      f[i][v] = max(f[i][v - 1] + (ai[v] == 0), f[i][v]);
      if (ai[v] == 1) g[i][v][1] = 1;
      g[i][v][1] = max(g[i][v][1], g[i][v - 1][1] + (ai[v] == 1));
      g[i][v][0] = max(g[i][v - 1][1], g[i][v - 1][0]) + (ai[v] == 0);
    }
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0][0] = dp[i - 1][0][0] + (ai[i] == 0);
    dp[i][0][1] = max(dp[i - 1][0][1], dp[i - 1][0][0]) + (ai[i] == 1);
    dp[i][1][1] = max(dp[i - 1][0][1], dp[i - 1][0][0]) + (ai[i] == 1);
    dp[i][1][1] =
        max(max(dp[i - 1][1][1], dp[i - 1][1][0]) + (ai[i] == 1), dp[i][1][1]);
    dp[i][1][0] = dp[i - 1][0][0] + (ai[i] == 0);
    dp[i][1][0] = max(dp[i - 1][1][0] + (ai[i] == 0), dp[i][1][0]);
    for (int v = i; v >= 1; --v) {
      dp[i][1][0] = max(dp[v - 1][0][0] + f[v][i], dp[i][1][0]);
      dp[i][1][1] =
          max(dp[v - 1][0][0] + max(g[v][i][0], g[v][i][1]), dp[i][1][1]);
      dp[i][1][1] = max(dp[v - 1][0][1] + g[v][i][1], dp[i][1][1]);
    }
  }
  int ans = max(max(dp[n][1][0], dp[n][1][1]), max(dp[n][0][0], dp[n][0][1]));
  std::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 maxn = 2e3 + 5;
const int INF = -0x3f3f3f3f;
int n, ai[maxn], f[maxn][maxn], g[maxn][maxn][2], dp[maxn][2][2];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ai[i]);
    ai[i]--;
  }
  memset(g, -0x3f, sizeof(g));
  memset(f, -0x3f, sizeof(f));
  for (int i = 1; i <= n; i++) {
    if (ai[i] == 1)
      g[i][i][1] = 1;
    else
      f[i][i] = 1;
    for (int v = i + 1; v <= n; v++) {
      if (ai[v] == 0) f[i][v] = 1;
      f[i][v] = max(f[i][v - 1] + (ai[v] == 0), f[i][v]);
      if (ai[v] == 1) g[i][v][1] = 1;
      g[i][v][1] = max(g[i][v][1], g[i][v - 1][1] + (ai[v] == 1));
      g[i][v][0] = max(g[i][v - 1][1], g[i][v - 1][0]) + (ai[v] == 0);
    }
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0][0] = dp[i - 1][0][0] + (ai[i] == 0);
    dp[i][0][1] = max(dp[i - 1][0][1], dp[i - 1][0][0]) + (ai[i] == 1);
    dp[i][1][1] = max(dp[i - 1][0][1], dp[i - 1][0][0]) + (ai[i] == 1);
    dp[i][1][1] =
        max(max(dp[i - 1][1][1], dp[i - 1][1][0]) + (ai[i] == 1), dp[i][1][1]);
    dp[i][1][0] = dp[i - 1][0][0] + (ai[i] == 0);
    dp[i][1][0] = max(dp[i - 1][1][0] + (ai[i] == 0), dp[i][1][0]);
    for (int v = i; v >= 1; --v) {
      dp[i][1][0] = max(dp[v - 1][0][0] + f[v][i], dp[i][1][0]);
      dp[i][1][1] =
          max(dp[v - 1][0][0] + max(g[v][i][0], g[v][i][1]), dp[i][1][1]);
      dp[i][1][1] = max(dp[v - 1][0][1] + g[v][i][1], dp[i][1][1]);
    }
  }
  int ans = max(max(dp[n][1][0], dp[n][1][1]), max(dp[n][0][0], dp[n][0][1]));
  std::cout << ans;
  return 0;
}

```