#include <bits/stdc++.h>
using namespace std;
long long arr[1000000], dp[1000000], dp1[100000], dp2[100000], dpp1[100000],
    dpp2[100000], seg[1000000];
long long build(long long in, long long l, long long r) {
  if (l == r) {
    seg[in] = l;
    return dp[l];
  }
  long long mid = (l + r) / 2;
  if (build(in * 2 + 1, l, mid) < build(in * 2 + 2, mid + 1, r)) {
    seg[in] = seg[in * 2 + 2];
    return dp[seg[in]];
  } else {
    seg[in] = seg[in * 2 + 1];
    return dp[seg[in]];
  }
}
long long query(long long in, long long l, long long r, long long ql,
                long long qr) {
  if (l > qr || r < ql) {
    return -1;
  }
  if (l >= ql && r <= qr) {
    return seg[in];
  }
  long long mid = (l + r) / 2;
  long long rl = query(in * 2 + 1, l, mid, ql, qr),
            rr = query(in * 2 + 2, mid + 1, r, ql, qr);
  if (rl == -1) return rr;
  if (rr == -1) return rl;
  if (dp[rl] < dp[rr]) return rr;
  return rl;
}
int main() {
  ios::sync_with_stdio(0);
  long long n, c2 = 0, c1 = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; i++) {
    if (arr[i] == 2)
      c2++;
    else
      c1++;
    dp[i] = c2;
    dp2[i] = c2;
    dpp1[i] = c1;
  }
  c2 = c1 = 0;
  for (int i = n - 1; i >= 0; i--) {
    dp[i] += c1;
    if (arr[i] == 1)
      c1++;
    else
      c2++;
    dp1[i] = c1;
    dpp2[i] = c2;
  }
  build(0, 0, n - 1);
  long long ans = dpp2[0];
  for (int i = 0; i < n; i++) ans = max(ans, dpp1[i] + dpp2[i + 1]);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (i != 0)
        ans = max(ans, dp[query(0, 0, n - 1, i, j)] - dp2[i - 1] - dp1[j + 1] +
                           dpp1[i - 1] + dpp2[j + 1]);
      else
        ans = max(ans, dp[query(0, 0, n - 1, i, j)] - dp1[j + 1] + dpp2[j + 1]);
    }
  }
  cout << ans;
}

### Prompt

Your task is to create 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;
long long arr[1000000], dp[1000000], dp1[100000], dp2[100000], dpp1[100000],
    dpp2[100000], seg[1000000];
long long build(long long in, long long l, long long r) {
  if (l == r) {
    seg[in] = l;
    return dp[l];
  }
  long long mid = (l + r) / 2;
  if (build(in * 2 + 1, l, mid) < build(in * 2 + 2, mid + 1, r)) {
    seg[in] = seg[in * 2 + 2];
    return dp[seg[in]];
  } else {
    seg[in] = seg[in * 2 + 1];
    return dp[seg[in]];
  }
}
long long query(long long in, long long l, long long r, long long ql,
                long long qr) {
  if (l > qr || r < ql) {
    return -1;
  }
  if (l >= ql && r <= qr) {
    return seg[in];
  }
  long long mid = (l + r) / 2;
  long long rl = query(in * 2 + 1, l, mid, ql, qr),
            rr = query(in * 2 + 2, mid + 1, r, ql, qr);
  if (rl == -1) return rr;
  if (rr == -1) return rl;
  if (dp[rl] < dp[rr]) return rr;
  return rl;
}
int main() {
  ios::sync_with_stdio(0);
  long long n, c2 = 0, c1 = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; i++) {
    if (arr[i] == 2)
      c2++;
    else
      c1++;
    dp[i] = c2;
    dp2[i] = c2;
    dpp1[i] = c1;
  }
  c2 = c1 = 0;
  for (int i = n - 1; i >= 0; i--) {
    dp[i] += c1;
    if (arr[i] == 1)
      c1++;
    else
      c2++;
    dp1[i] = c1;
    dpp2[i] = c2;
  }
  build(0, 0, n - 1);
  long long ans = dpp2[0];
  for (int i = 0; i < n; i++) ans = max(ans, dpp1[i] + dpp2[i + 1]);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (i != 0)
        ans = max(ans, dp[query(0, 0, n - 1, i, j)] - dp2[i - 1] - dp1[j + 1] +
                           dpp1[i - 1] + dpp2[j + 1]);
      else
        ans = max(ans, dp[query(0, 0, n - 1, i, j)] - dp1[j + 1] + dpp2[j + 1]);
    }
  }
  cout << ans;
}

```