#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2003;
int pre[MAXN][2];
int suf[MAXN][2];
int g[MAXN][MAXN][2][2];
int N;
int w[MAXN];
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &w[i]);
    w[i]--;
  }
  for (int i = 0; i < N; i++) {
    if (i > 0) memcpy(pre[i], pre[i - 1], sizeof(pre[i]));
    pre[i][w[i]]++;
  }
  for (int j = N - 1; j > 0; j--) {
    if (j < N - 1) memcpy(suf[j], suf[j + 1], sizeof(suf[j]));
    suf[j][w[j]]++;
  }
  for (int i = 0; i < N; i++)
    for (int k = 0; k < w[i] + 1; k++) g[i][i][w[i]][k] = 1;
  for (int i = 0; i < N; i++)
    for (int j = i + 1; j < N; j++) {
      g[i][j][0][0] = g[i][j - 1][0][0] + g[j][j][0][0];
      g[i][j][1][0] = max(g[i][j - 1][1][1] + g[j][j][1][0],
                          g[i][j - 1][1][0] + g[j][j][0][0]);
      g[i][j][1][1] = g[i][j - 1][1][1] + g[j][j][1][1];
    }
  int ans = max(pre[N - 1][1], pre[N - 1][0]);
  ans = max(ans, g[0][N - 1][1][0]);
  for (int i = 0; i < N; i++) {
    for (int j = i; j < N; j++) {
      if (i > 0 && j < N - 1)
        ans = max(ans, pre[i - 1][0] + g[i][j][1][0] + suf[j + 1][1]);
    }
  }
  for (int i = 0; i < N; i++) {
    if (i != N - 1) ans = max(ans, g[0][i][1][0] + suf[i + 1][1]);
    if (i) ans = max(ans, pre[i - 1][0] + g[i][N - 1][1][0]);
  }
  printf("%d", 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;
const int MAXN = 2003;
int pre[MAXN][2];
int suf[MAXN][2];
int g[MAXN][MAXN][2][2];
int N;
int w[MAXN];
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &w[i]);
    w[i]--;
  }
  for (int i = 0; i < N; i++) {
    if (i > 0) memcpy(pre[i], pre[i - 1], sizeof(pre[i]));
    pre[i][w[i]]++;
  }
  for (int j = N - 1; j > 0; j--) {
    if (j < N - 1) memcpy(suf[j], suf[j + 1], sizeof(suf[j]));
    suf[j][w[j]]++;
  }
  for (int i = 0; i < N; i++)
    for (int k = 0; k < w[i] + 1; k++) g[i][i][w[i]][k] = 1;
  for (int i = 0; i < N; i++)
    for (int j = i + 1; j < N; j++) {
      g[i][j][0][0] = g[i][j - 1][0][0] + g[j][j][0][0];
      g[i][j][1][0] = max(g[i][j - 1][1][1] + g[j][j][1][0],
                          g[i][j - 1][1][0] + g[j][j][0][0]);
      g[i][j][1][1] = g[i][j - 1][1][1] + g[j][j][1][1];
    }
  int ans = max(pre[N - 1][1], pre[N - 1][0]);
  ans = max(ans, g[0][N - 1][1][0]);
  for (int i = 0; i < N; i++) {
    for (int j = i; j < N; j++) {
      if (i > 0 && j < N - 1)
        ans = max(ans, pre[i - 1][0] + g[i][j][1][0] + suf[j + 1][1]);
    }
  }
  for (int i = 0; i < N; i++) {
    if (i != N - 1) ans = max(ans, g[0][i][1][0] + suf[i + 1][1]);
    if (i) ans = max(ans, pre[i - 1][0] + g[i][N - 1][1][0]);
  }
  printf("%d", ans);
  return 0;
}

```