#include <bits/stdc++.h>
using namespace std;
int one[2005], a[2005], is[2005], ds[2005];
pair<int, int> b[2005];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) {
    return a.second > b.second;
  }
  return a.first < b.first;
}
int tr[2005][4 * 2005];
int get(int c, int n, int b, int e, int i, int j) {
  if (i > j) return 0;
  if (e < i || j < b) return 0;
  if (i <= b && j >= e) return tr[c][n];
  int l, r, m;
  m = (b + e) >> 1;
  l = n << 1;
  r = l + 1;
  return max(get(c, l, b, m, i, j), get(c, r, m + 1, e, i, j));
}
void update(int c, int n, int b, int e, int i, int x) {
  if (i < b || i > e) return;
  if (b == e) {
    tr[c][n] = x;
    return;
  }
  int l, r, m;
  m = (b + e) >> 1;
  l = n << 1;
  r = l + 1;
  update(c, l, b, m, i, x);
  update(c, r, m + 1, e, i, x);
  tr[c][n] = max(tr[c][l], tr[c][r]);
}
int main() {
  int n, ans = 0, g;
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i = i + 1) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      one[i] = one[i - 1] + 1;
    else
      one[i] = one[i - 1];
  }
  for (int i = 1; i < n + 1; i = i + 1) {
    is[i] = 1;
    for (int j = 1; j < i; j = j + 1) {
      if (a[j] <= a[i]) is[i] = max(is[i], is[j] + 1);
    }
    ans = max(ans, is[i]);
  }
  for (int i = n; i > 0; i = i - 1) {
    ds[i] = 1;
    for (int j = n; j > i; j = j - 1) {
      if (a[j] >= a[i]) ds[i] = max(ds[i], ds[j] + 1);
    }
  }
  for (int i = n; i > 0; i = i - 1) {
    for (int j = i; j > 0; j = j - 1) b[j] = make_pair(a[j], j);
    sort(b + 1, b + i + 1, cmp);
    for (int j = 1; j < i + 1; j = j + 1) {
      g = get(i, 1, 1, i, b[j].second + 1, i);
      update(i, 1, 1, i, b[j].second, g + 1);
    }
  }
  for (int i = 1; i < n + 1; i = i + 1) {
    for (int j = i; j < n + 1; j = j + 1) {
      if ((one[j] - one[i - 1]) == (j - i + 1)) {
        ans = max(ans, j - i + 1 + one[i - 1] + ds[j + 1]);
      } else if ((one[j] - one[i - 1]) == 0) {
        ans = max(ans, is[i - 1] + (j - i + 1) + (n - j - one[n] + one[j]));
      } else {
        ans = max(ans, one[i - 1] + (n - j - one[n] + one[j]) +
                           get(j, 1, 1, j, i, j));
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
int one[2005], a[2005], is[2005], ds[2005];
pair<int, int> b[2005];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) {
    return a.second > b.second;
  }
  return a.first < b.first;
}
int tr[2005][4 * 2005];
int get(int c, int n, int b, int e, int i, int j) {
  if (i > j) return 0;
  if (e < i || j < b) return 0;
  if (i <= b && j >= e) return tr[c][n];
  int l, r, m;
  m = (b + e) >> 1;
  l = n << 1;
  r = l + 1;
  return max(get(c, l, b, m, i, j), get(c, r, m + 1, e, i, j));
}
void update(int c, int n, int b, int e, int i, int x) {
  if (i < b || i > e) return;
  if (b == e) {
    tr[c][n] = x;
    return;
  }
  int l, r, m;
  m = (b + e) >> 1;
  l = n << 1;
  r = l + 1;
  update(c, l, b, m, i, x);
  update(c, r, m + 1, e, i, x);
  tr[c][n] = max(tr[c][l], tr[c][r]);
}
int main() {
  int n, ans = 0, g;
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i = i + 1) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      one[i] = one[i - 1] + 1;
    else
      one[i] = one[i - 1];
  }
  for (int i = 1; i < n + 1; i = i + 1) {
    is[i] = 1;
    for (int j = 1; j < i; j = j + 1) {
      if (a[j] <= a[i]) is[i] = max(is[i], is[j] + 1);
    }
    ans = max(ans, is[i]);
  }
  for (int i = n; i > 0; i = i - 1) {
    ds[i] = 1;
    for (int j = n; j > i; j = j - 1) {
      if (a[j] >= a[i]) ds[i] = max(ds[i], ds[j] + 1);
    }
  }
  for (int i = n; i > 0; i = i - 1) {
    for (int j = i; j > 0; j = j - 1) b[j] = make_pair(a[j], j);
    sort(b + 1, b + i + 1, cmp);
    for (int j = 1; j < i + 1; j = j + 1) {
      g = get(i, 1, 1, i, b[j].second + 1, i);
      update(i, 1, 1, i, b[j].second, g + 1);
    }
  }
  for (int i = 1; i < n + 1; i = i + 1) {
    for (int j = i; j < n + 1; j = j + 1) {
      if ((one[j] - one[i - 1]) == (j - i + 1)) {
        ans = max(ans, j - i + 1 + one[i - 1] + ds[j + 1]);
      } else if ((one[j] - one[i - 1]) == 0) {
        ans = max(ans, is[i - 1] + (j - i + 1) + (n - j - one[n] + one[j]));
      } else {
        ans = max(ans, one[i - 1] + (n - j - one[n] + one[j]) +
                           get(j, 1, 1, j, i, j));
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```