#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("-O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int INF = 2e9;
const int64_t INFL = 9e18;
const bool DEBUG_ = 0;
const double EPS = 1e-6;
const int MOD7 = 1e9 + 7;
const int MOD9 = 1e9 + 9;
const int MOD3 = 998244353;
int main() {
  int n;
  cin >> n;
  vector<int> a(1 + n);
  for (int i = 1; i < 1 + n; i++) {
    cin >> a[i];
  }
  vector<vector<int>> cntf(3, vector<int>(2 + n, 0));
  vector<vector<int>> cntr(3, vector<int>(2 + n, 0));
  for (int i = 1; i < 1 + n; i++) {
    cntf[1][i] = cntf[1][i - 1] + (a[i] == 1);
    cntf[2][i] = cntf[2][i - 1] + (a[i] == 2);
  }
  for (int i = n; i >= 1; i--) {
    cntr[1][i] = cntr[1][i + 1] + (a[i] == 1);
    cntr[2][i] = cntr[2][i + 1] + (a[i] == 2);
  }
  function<int(int, int)> getmf = [&](int l, int r) {
    int res = -INF;
    for (int i = l; i < r + 1; i++) {
      int cnt1 = cntf[1][i] - cntf[1][l - 1];
      int cnt2 = cntr[2][i] - cntr[2][r + 1];
      if (cnt1 + cnt2 > res) {
        res = cnt1 + cnt2;
      }
    }
    return res;
  };
  function<int(int, int)> getmr = [&](int l, int r) {
    int res = -INF;
    for (int i = r; i >= l; i--) {
      int cnt1 = cntr[1][i] - cntr[1][r + 1];
      int cnt2 = cntf[2][i] - cntf[2][l - 1];
      if (cnt1 + cnt2 > res) {
        res = cnt1 + cnt2;
      }
    }
    return res;
  };
  int maxv = -INF, lm = 0, rm = 0;
  for (int l = 1; l < 1 + n; l++) {
    for (int r = l; r < 1 + n; r++) {
      int cur = getmr(l, r) - getmf(l, r);
      if (cur > maxv) {
        maxv = cur;
        lm = l;
        rm = r;
      }
    }
  }
  cout << getmf(1, n) + (maxv > 0 ? maxv : 0);
  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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("-O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int INF = 2e9;
const int64_t INFL = 9e18;
const bool DEBUG_ = 0;
const double EPS = 1e-6;
const int MOD7 = 1e9 + 7;
const int MOD9 = 1e9 + 9;
const int MOD3 = 998244353;
int main() {
  int n;
  cin >> n;
  vector<int> a(1 + n);
  for (int i = 1; i < 1 + n; i++) {
    cin >> a[i];
  }
  vector<vector<int>> cntf(3, vector<int>(2 + n, 0));
  vector<vector<int>> cntr(3, vector<int>(2 + n, 0));
  for (int i = 1; i < 1 + n; i++) {
    cntf[1][i] = cntf[1][i - 1] + (a[i] == 1);
    cntf[2][i] = cntf[2][i - 1] + (a[i] == 2);
  }
  for (int i = n; i >= 1; i--) {
    cntr[1][i] = cntr[1][i + 1] + (a[i] == 1);
    cntr[2][i] = cntr[2][i + 1] + (a[i] == 2);
  }
  function<int(int, int)> getmf = [&](int l, int r) {
    int res = -INF;
    for (int i = l; i < r + 1; i++) {
      int cnt1 = cntf[1][i] - cntf[1][l - 1];
      int cnt2 = cntr[2][i] - cntr[2][r + 1];
      if (cnt1 + cnt2 > res) {
        res = cnt1 + cnt2;
      }
    }
    return res;
  };
  function<int(int, int)> getmr = [&](int l, int r) {
    int res = -INF;
    for (int i = r; i >= l; i--) {
      int cnt1 = cntr[1][i] - cntr[1][r + 1];
      int cnt2 = cntf[2][i] - cntf[2][l - 1];
      if (cnt1 + cnt2 > res) {
        res = cnt1 + cnt2;
      }
    }
    return res;
  };
  int maxv = -INF, lm = 0, rm = 0;
  for (int l = 1; l < 1 + n; l++) {
    for (int r = l; r < 1 + n; r++) {
      int cur = getmr(l, r) - getmf(l, r);
      if (cur > maxv) {
        maxv = cur;
        lm = l;
        rm = r;
      }
    }
  }
  cout << getmf(1, n) + (maxv > 0 ? maxv : 0);
  return 0;
}

```