#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxN = 2010;
int n;
int a[maxN];
int dp[maxN][maxN][2][2];
int pref[maxN][2], suff[maxN][2];
void getDp(int ind) {
  dp[ind][ind][a[ind]][a[ind]] = 1;
  for (int i = ind + 1; i <= n; i++) {
    if (a[i] == 0) {
      dp[ind][i][0][0] = dp[ind][i - 1][0][0] + 1;
      dp[ind][i][1][0] = max(dp[ind][i - 1][1][0], dp[ind][i - 1][1][1]) + 1;
      dp[ind][i][1][1] = dp[ind][i - 1][1][1];
    } else {
      dp[ind][i][0][0] = dp[ind][i - 1][0][0];
      dp[ind][i][1][0] = dp[ind][i - 1][1][0];
      dp[ind][i][1][1] = dp[ind][i - 1][1][1] + 1;
    }
  }
}
signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 1; i <= n; i++) getDp(i);
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0) {
      pref[i][0] = pref[i - 1][0] + 1;
      pref[i][1] = pref[i - 1][1];
    } else {
      pref[i][0] = pref[i - 1][0];
      pref[i][1] = max(pref[i - 1][0], pref[i - 1][1]) + 1;
    }
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] == 0) {
      suff[i][0] = max(suff[i + 1][0], suff[i + 1][1]) + 1;
      suff[i][1] = suff[i + 1][1];
    } else {
      suff[i][0] = suff[i + 1][0];
      suff[i][1] = suff[i + 1][1] + 1;
    }
  }
  int ans = max(pref[n][0], pref[n][1]);
  for (int l = 1; l < n; l++) {
    for (int r = l + 1; r <= n; r++) {
      ans = max(ans, pref[l - 1][0] + dp[l][r][1][0] + suff[r + 1][1]);
      ans = max(ans, pref[l - 1][0] + dp[l][r][0][0] + suff[r + 1][0]);
      ans = max(ans, pref[l - 1][0] + dp[l][r][0][0] + suff[r + 1][1]);
      ans = max(ans, pref[l - 1][0] + dp[l][r][1][1] + suff[r + 1][1]);
      ans = max(ans, pref[l - 1][1] + dp[l][r][1][1] + suff[r + 1][1]);
    }
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxN = 2010;
int n;
int a[maxN];
int dp[maxN][maxN][2][2];
int pref[maxN][2], suff[maxN][2];
void getDp(int ind) {
  dp[ind][ind][a[ind]][a[ind]] = 1;
  for (int i = ind + 1; i <= n; i++) {
    if (a[i] == 0) {
      dp[ind][i][0][0] = dp[ind][i - 1][0][0] + 1;
      dp[ind][i][1][0] = max(dp[ind][i - 1][1][0], dp[ind][i - 1][1][1]) + 1;
      dp[ind][i][1][1] = dp[ind][i - 1][1][1];
    } else {
      dp[ind][i][0][0] = dp[ind][i - 1][0][0];
      dp[ind][i][1][0] = dp[ind][i - 1][1][0];
      dp[ind][i][1][1] = dp[ind][i - 1][1][1] + 1;
    }
  }
}
signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 1; i <= n; i++) getDp(i);
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0) {
      pref[i][0] = pref[i - 1][0] + 1;
      pref[i][1] = pref[i - 1][1];
    } else {
      pref[i][0] = pref[i - 1][0];
      pref[i][1] = max(pref[i - 1][0], pref[i - 1][1]) + 1;
    }
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] == 0) {
      suff[i][0] = max(suff[i + 1][0], suff[i + 1][1]) + 1;
      suff[i][1] = suff[i + 1][1];
    } else {
      suff[i][0] = suff[i + 1][0];
      suff[i][1] = suff[i + 1][1] + 1;
    }
  }
  int ans = max(pref[n][0], pref[n][1]);
  for (int l = 1; l < n; l++) {
    for (int r = l + 1; r <= n; r++) {
      ans = max(ans, pref[l - 1][0] + dp[l][r][1][0] + suff[r + 1][1]);
      ans = max(ans, pref[l - 1][0] + dp[l][r][0][0] + suff[r + 1][0]);
      ans = max(ans, pref[l - 1][0] + dp[l][r][0][0] + suff[r + 1][1]);
      ans = max(ans, pref[l - 1][0] + dp[l][r][1][1] + suff[r + 1][1]);
      ans = max(ans, pref[l - 1][1] + dp[l][r][1][1] + suff[r + 1][1]);
    }
  }
  cout << ans << "\n";
  return 0;
}

```