#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
const int maxn = 2e3 + 20;
const int maxm = 2e3 + 20;
int in(int &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return -1;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
int f1[maxn];
int f2[maxn];
int h1[maxn];
int h2[maxn];
int g1[maxn], g2[maxn];
int a[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    if (a[i] & 1) {
      f1[i] = f1[i - 1] + 1;
      f2[i] = f2[i - 1];
    } else {
      f1[i] = f1[i - 1];
      f2[i] = max(f2[i - 1], f1[i - 1]) + 1;
    }
  }
  for (int i = n; i >= 1; --i) {
    if (a[i] & 1) {
      g2[i] = g2[i + 1];
      g1[i] = max(g2[i + 1], g1[i + 1]) + 1;
    } else {
      g2[i] = g2[i + 1] + 1;
      g1[i] = g1[i + 1];
    }
  }
  int ans = 1;
  for (int i = 1; i <= n; ++i) {
    h1[i - 1] = h2[i - 1] = 0;
    int cnt1 = 0;
    for (int j = i; j <= n; ++j) {
      if (a[j] & 1) {
        ++cnt1;
        h2[j] = h2[j - 1];
        h1[j] = max(h2[j - 1], h1[j - 1]) + 1;
      } else {
        h2[j] = h2[j - 1] + 1;
        h1[j] = h1[j - 1];
      }
      int q = f1[i - 1] + cnt1 + g1[j + 1];
      ans = max(ans, q);
      q = f1[i - 1] + h1[j] + g2[j + 1];
      ans = max(ans, q);
      q = f1[i - 1] + h2[j] + g2[j + 1];
      ans = max(ans, q);
      q = f2[i - 1] + h2[j] + g2[j + 1];
      ans = max(ans, q);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 long long mod = 1000000007;
const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
const int maxn = 2e3 + 20;
const int maxm = 2e3 + 20;
int in(int &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return -1;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
int f1[maxn];
int f2[maxn];
int h1[maxn];
int h2[maxn];
int g1[maxn], g2[maxn];
int a[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    if (a[i] & 1) {
      f1[i] = f1[i - 1] + 1;
      f2[i] = f2[i - 1];
    } else {
      f1[i] = f1[i - 1];
      f2[i] = max(f2[i - 1], f1[i - 1]) + 1;
    }
  }
  for (int i = n; i >= 1; --i) {
    if (a[i] & 1) {
      g2[i] = g2[i + 1];
      g1[i] = max(g2[i + 1], g1[i + 1]) + 1;
    } else {
      g2[i] = g2[i + 1] + 1;
      g1[i] = g1[i + 1];
    }
  }
  int ans = 1;
  for (int i = 1; i <= n; ++i) {
    h1[i - 1] = h2[i - 1] = 0;
    int cnt1 = 0;
    for (int j = i; j <= n; ++j) {
      if (a[j] & 1) {
        ++cnt1;
        h2[j] = h2[j - 1];
        h1[j] = max(h2[j - 1], h1[j - 1]) + 1;
      } else {
        h2[j] = h2[j - 1] + 1;
        h1[j] = h1[j - 1];
      }
      int q = f1[i - 1] + cnt1 + g1[j + 1];
      ans = max(ans, q);
      q = f1[i - 1] + h1[j] + g2[j + 1];
      ans = max(ans, q);
      q = f1[i - 1] + h2[j] + g2[j + 1];
      ans = max(ans, q);
      q = f2[i - 1] + h2[j] + g2[j + 1];
      ans = max(ans, q);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```