#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 block_size = 320;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 7;
const long double eps = 1e-9;
const double PI = atan(1) * 4;
template <typename T>
inline int sign(const T &a) {
  if (a < 0) return -1;
  if (a > 0) return 1;
  return 0;
}
template <typename T, typename S>
inline bool upmin(T &a, const S &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T, typename S>
inline bool upmax(T &a, const S &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline void in(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
long long twop(int x) { return 1LL << x; }
template <typename A, typename B>
inline void in(A &x, B &y) {
  in(x);
  in(y);
}
template <typename A, typename B, typename C>
inline void in(A &x, B &y, C &z) {
  in(x);
  in(y);
  in(z);
}
template <typename A, typename B, typename C, typename D>
inline void in(A &x, B &y, C &z, D &d) {
  in(x);
  in(y);
  in(z);
  in(d);
}
namespace SOLVE {
long long n;
long long a[100000];
long long dp[2];
long long pre[20000][2];
long long getSum(long long l, long long r, long long v) {
  return pre[r][v] - pre[l - 1][v];
}
void main() {
  in(n);
  n++;
  for (long long i = 1; i < n; i++) {
    in(a[i]);
    a[i]--;
    pre[i][a[i]] = 1;
  }
  for (long long i = 1; i < n; i++) {
    for (long long j = 0; j < 2; j++) pre[i][j] += pre[i - 1][j];
  }
  memset(dp, 0xcf, sizeof(dp));
  for (long long i = 1; i < n; i++) {
    if (a[i] == 0) {
      dp[0]++;
    } else {
      dp[1]++;
      upmax(dp[1], dp[0] + 1);
    }
  }
  long long ans = max(dp[0], dp[1]);
  for (long long i = 1; i < n; i++) {
    memset(dp, 0, sizeof(dp));
    for (long long j = i; j < n; j++) {
      if (a[j] == 1) {
        dp[1]++;
      } else {
        dp[0]++;
        upmax(dp[0], dp[1] + 1);
      }
      upmax(ans,
            getSum(1, i - 1, 0) + getSum(j + 1, n - 1, 1) + max(dp[0], dp[1]));
    }
  }
  cout << ans << endl;
}
}  // namespace SOLVE
int main() {
  SOLVE::main();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 block_size = 320;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 7;
const long double eps = 1e-9;
const double PI = atan(1) * 4;
template <typename T>
inline int sign(const T &a) {
  if (a < 0) return -1;
  if (a > 0) return 1;
  return 0;
}
template <typename T, typename S>
inline bool upmin(T &a, const S &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T, typename S>
inline bool upmax(T &a, const S &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline void in(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
long long twop(int x) { return 1LL << x; }
template <typename A, typename B>
inline void in(A &x, B &y) {
  in(x);
  in(y);
}
template <typename A, typename B, typename C>
inline void in(A &x, B &y, C &z) {
  in(x);
  in(y);
  in(z);
}
template <typename A, typename B, typename C, typename D>
inline void in(A &x, B &y, C &z, D &d) {
  in(x);
  in(y);
  in(z);
  in(d);
}
namespace SOLVE {
long long n;
long long a[100000];
long long dp[2];
long long pre[20000][2];
long long getSum(long long l, long long r, long long v) {
  return pre[r][v] - pre[l - 1][v];
}
void main() {
  in(n);
  n++;
  for (long long i = 1; i < n; i++) {
    in(a[i]);
    a[i]--;
    pre[i][a[i]] = 1;
  }
  for (long long i = 1; i < n; i++) {
    for (long long j = 0; j < 2; j++) pre[i][j] += pre[i - 1][j];
  }
  memset(dp, 0xcf, sizeof(dp));
  for (long long i = 1; i < n; i++) {
    if (a[i] == 0) {
      dp[0]++;
    } else {
      dp[1]++;
      upmax(dp[1], dp[0] + 1);
    }
  }
  long long ans = max(dp[0], dp[1]);
  for (long long i = 1; i < n; i++) {
    memset(dp, 0, sizeof(dp));
    for (long long j = i; j < n; j++) {
      if (a[j] == 1) {
        dp[1]++;
      } else {
        dp[0]++;
        upmax(dp[0], dp[1] + 1);
      }
      upmax(ans,
            getSum(1, i - 1, 0) + getSum(j + 1, n - 1, 1) + max(dp[0], dp[1]));
    }
  }
  cout << ans << endl;
}
}  // namespace SOLVE
int main() {
  SOLVE::main();
  return 0;
}

```