#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7, MAXN = 1e5 + 10;
int n;
int p[MAXN], ll[MAXN], a[MAXN], s[MAXN], rr[MAXN], ans, k, mn;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    p[i] = p[i - 1];
    if (a[i] == 1)
      p[i]--;
    else
      p[i]++;
    mn = min(mn, p[i]);
    ll[i] = p[i] - mn;
  }
  for (int i = n; i >= 1; --i) {
    s[i] = s[i + 1];
    if (a[i] == 1)
      ++s[i];
    else
      --s[i];
    k = min(k, s[i]);
    rr[i] = s[i] - k;
  }
  for (int i = 1; i <= n + 1; ++i) {
    int m = 0;
    for (int j = 1; j < i; ++j) m += (a[j] == 1);
    for (int j = i; j <= n; ++j) m += (a[j] == 2);
    int res = 0;
    for (int j = i; j <= n; ++j) res = max(res, rr[j]);
    for (int j = 1; j < i; ++j) res = max(res, ll[j]);
    int l = i - 1, r;
    while (l && a[l] == 2) --l;
    r = i;
    while (r <= n && a[r] == 1) ++r;
    res = max(res, r - l - 1);
    ans = max(ans, m + res);
  }
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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 INF = 1e9 + 7, MAXN = 1e5 + 10;
int n;
int p[MAXN], ll[MAXN], a[MAXN], s[MAXN], rr[MAXN], ans, k, mn;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    p[i] = p[i - 1];
    if (a[i] == 1)
      p[i]--;
    else
      p[i]++;
    mn = min(mn, p[i]);
    ll[i] = p[i] - mn;
  }
  for (int i = n; i >= 1; --i) {
    s[i] = s[i + 1];
    if (a[i] == 1)
      ++s[i];
    else
      --s[i];
    k = min(k, s[i]);
    rr[i] = s[i] - k;
  }
  for (int i = 1; i <= n + 1; ++i) {
    int m = 0;
    for (int j = 1; j < i; ++j) m += (a[j] == 1);
    for (int j = i; j <= n; ++j) m += (a[j] == 2);
    int res = 0;
    for (int j = i; j <= n; ++j) res = max(res, rr[j]);
    for (int j = 1; j < i; ++j) res = max(res, ll[j]);
    int l = i - 1, r;
    while (l && a[l] == 2) --l;
    r = i;
    while (r <= n && a[r] == 1) ++r;
    res = max(res, r - l - 1);
    ans = max(ans, m + res);
  }
  cout << ans;
  return 0;
}

```