#include <bits/stdc++.h>
using namespace std;
struct IO {
  char buf[(1 << 20)], *p1, *p2;
  char pbuf[(1 << 20)], *pp;
  IO() : p1(buf), p2(buf), pp(pbuf) {}
  inline char gc() {
    return getchar();
    if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, (1 << 20), stdin);
    return p1 == p2 ? ' ' : *p1++;
  }
  inline bool blank(char ch) {
    return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
  }
  template <class T>
  inline void read(T &x) {
    register double tmp = 1;
    register bool sign = 0;
    x = 0;
    register char ch = gc();
    for (; !(ch >= '0' && ch <= '9'); ch = gc())
      if (ch == '-') sign = 1;
    for (; (ch >= '0' && ch <= '9'); ch = gc()) x = x * 10 + (ch - '0');
    if (ch == '.')
      for (ch = gc(); (ch >= '0' && ch <= '9'); ch = gc())
        tmp /= 10.0, x += tmp * (ch - '0');
    if (sign) x = -x;
  }
  inline void read(char *s) {
    register char ch = gc();
    for (; blank(ch); ch = gc())
      ;
    for (; !blank(ch); ch = gc()) *s++ = ch;
    *s = 0;
  }
  inline void read(char &c) {
    for (c = gc(); blank(c); c = gc())
      ;
  }
  template <class t>
  inline void write(t x) {
    if (x < 0)
      putchar('-'), write(-x);
    else {
      if (x > 9) write(x / 10);
      putchar('0' + x % 10);
    }
  }
} io;
const int mod = 1e9 + 7;
const int mo = 998244353;
const int N = 5e5 + 5;
int n, m, s1, s0, f[N][4], a[N];
int main() {
  io.read(n);
  for (int i = (1); i <= (n); i++) io.read(a[i]), a[i]--;
  for (int i = (1); i <= (n); i++) {
    int x = a[i];
    f[i][0] = f[i - 1][0] + (!a[i]);
    f[i][1] = max(f[i][0], f[i - 1][1] + (a[i]));
    f[i][2] = max(f[i][1], f[i - 1][2] + (!a[i]));
    f[i][3] = max(f[i][2], f[i - 1][3] + (a[i]));
  }
  io.write(f[n][3]);
  return 0;
}

### Prompt

Please formulate 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;
struct IO {
  char buf[(1 << 20)], *p1, *p2;
  char pbuf[(1 << 20)], *pp;
  IO() : p1(buf), p2(buf), pp(pbuf) {}
  inline char gc() {
    return getchar();
    if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, (1 << 20), stdin);
    return p1 == p2 ? ' ' : *p1++;
  }
  inline bool blank(char ch) {
    return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
  }
  template <class T>
  inline void read(T &x) {
    register double tmp = 1;
    register bool sign = 0;
    x = 0;
    register char ch = gc();
    for (; !(ch >= '0' && ch <= '9'); ch = gc())
      if (ch == '-') sign = 1;
    for (; (ch >= '0' && ch <= '9'); ch = gc()) x = x * 10 + (ch - '0');
    if (ch == '.')
      for (ch = gc(); (ch >= '0' && ch <= '9'); ch = gc())
        tmp /= 10.0, x += tmp * (ch - '0');
    if (sign) x = -x;
  }
  inline void read(char *s) {
    register char ch = gc();
    for (; blank(ch); ch = gc())
      ;
    for (; !blank(ch); ch = gc()) *s++ = ch;
    *s = 0;
  }
  inline void read(char &c) {
    for (c = gc(); blank(c); c = gc())
      ;
  }
  template <class t>
  inline void write(t x) {
    if (x < 0)
      putchar('-'), write(-x);
    else {
      if (x > 9) write(x / 10);
      putchar('0' + x % 10);
    }
  }
} io;
const int mod = 1e9 + 7;
const int mo = 998244353;
const int N = 5e5 + 5;
int n, m, s1, s0, f[N][4], a[N];
int main() {
  io.read(n);
  for (int i = (1); i <= (n); i++) io.read(a[i]), a[i]--;
  for (int i = (1); i <= (n); i++) {
    int x = a[i];
    f[i][0] = f[i - 1][0] + (!a[i]);
    f[i][1] = max(f[i][0], f[i - 1][1] + (a[i]));
    f[i][2] = max(f[i][1], f[i - 1][2] + (!a[i]));
    f[i][3] = max(f[i][2], f[i - 1][3] + (a[i]));
  }
  io.write(f[n][3]);
  return 0;
}

```