#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2007;
const int INF = 1e9 + 7;
int a[MAXN];
int pref[MAXN];
int pst[MAXN];
int dp[MAXN][MAXN];
int dp1[MAXN][MAXN];
signed main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      pref[i] = pref[i - 1] - 1;
    } else {
      pref[i] = pref[i - 1] + 1;
    }
  }
  for (int i = n; i >= 1; --i) {
    if (a[i] == 2) {
      pst[i] = pst[i + 1] - 1;
    } else {
      pst[i] = pst[i + 1] + 1;
    }
  }
  for (int i = 0; i < MAXN; ++i) {
    for (int j = 0; j < MAXN; ++j) {
      dp[i][j] = dp1[i][j] = -INF;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int len = 0; len <= i; ++len) {
      dp[i][len] = max(dp[i - 1][len], pref[i] - pref[i - len]);
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int len = 0; len <= n - i + 1; ++len) {
      dp1[i][len] = max(dp1[i + 1][len], pst[i] - pst[i + len]);
    }
  }
  int ans = -INF;
  for (int p = 0; p <= n; ++p) {
    int left = 0;
    for (int i = 1; i <= p; ++i) {
      left += (a[i] == 1);
    }
    int right = 0;
    for (int i = p + 1; i <= n; ++i) {
      right += (a[i] == 2);
    }
    ans = max(ans, left + right);
    int curr = 0;
    for (int len = 1; len <= n - p; ++len) {
      int pos = p + len;
      if (a[pos] == 2) {
        --curr;
      } else {
        ++curr;
      }
      ans = max(ans, left + right + curr + dp[p][len]);
    }
    curr = 0;
    for (int len = 1; len <= p; ++len) {
      int pos = p - len + 1;
      if (a[pos] == 1) {
        --curr;
      } else {
        ++curr;
      }
      ans = max(ans, left + right + curr + dp1[p + 1][len]);
    }
  }
  cout << ans << '\n';
  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 MAXN = 2007;
const int INF = 1e9 + 7;
int a[MAXN];
int pref[MAXN];
int pst[MAXN];
int dp[MAXN][MAXN];
int dp1[MAXN][MAXN];
signed main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      pref[i] = pref[i - 1] - 1;
    } else {
      pref[i] = pref[i - 1] + 1;
    }
  }
  for (int i = n; i >= 1; --i) {
    if (a[i] == 2) {
      pst[i] = pst[i + 1] - 1;
    } else {
      pst[i] = pst[i + 1] + 1;
    }
  }
  for (int i = 0; i < MAXN; ++i) {
    for (int j = 0; j < MAXN; ++j) {
      dp[i][j] = dp1[i][j] = -INF;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int len = 0; len <= i; ++len) {
      dp[i][len] = max(dp[i - 1][len], pref[i] - pref[i - len]);
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int len = 0; len <= n - i + 1; ++len) {
      dp1[i][len] = max(dp1[i + 1][len], pst[i] - pst[i + len]);
    }
  }
  int ans = -INF;
  for (int p = 0; p <= n; ++p) {
    int left = 0;
    for (int i = 1; i <= p; ++i) {
      left += (a[i] == 1);
    }
    int right = 0;
    for (int i = p + 1; i <= n; ++i) {
      right += (a[i] == 2);
    }
    ans = max(ans, left + right);
    int curr = 0;
    for (int len = 1; len <= n - p; ++len) {
      int pos = p + len;
      if (a[pos] == 2) {
        --curr;
      } else {
        ++curr;
      }
      ans = max(ans, left + right + curr + dp[p][len]);
    }
    curr = 0;
    for (int len = 1; len <= p; ++len) {
      int pos = p - len + 1;
      if (a[pos] == 1) {
        --curr;
      } else {
        ++curr;
      }
      ans = max(ans, left + right + curr + dp1[p + 1][len]);
    }
  }
  cout << ans << '\n';
  return 0;
}

```