#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmax(T& a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
bool chkmin(T& a, T b) {
  return a > b ? a = b, 1 : 0;
}
const int maxn = 2000 + 10;
int n, ans;
int a[maxn], cnt[maxn];
int s1[maxn], s2[maxn];
int f[maxn][maxn], cntf[maxn][maxn];
int g[maxn][maxn], cntg[maxn][maxn];
inline void read(int& x) {
  x = 0;
  int p = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - '0');
    c = getchar();
  }
  x *= p;
}
inline void file() {
  freopen("C.in", "r", stdin);
  freopen("C.out", "w", stdout);
}
inline void build_s() {
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      s1[i] = s1[i - 1] + 1;
      s2[i] = s2[i - 1];
    } else {
      s1[i] = s1[i - 1];
      s2[i] = s2[i - 1] + 1;
    }
  }
}
inline void build_f() {
  for (int i = 1; i <= n - 1; ++i) {
    f[i][i] = 1;
    memset(cnt, 0, sizeof(cnt));
    if (a[i] == 1) cnt[i] = 1;
    int maxl = 1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 2) {
        f[i][j] = max(maxl, cnt[j - 1]) + 1;
        cnt[j] = cnt[j - 1];
        chkmax(maxl, f[i][j]);
      } else {
        f[i][j] = cnt[j - 1] + 1;
        cnt[j] = cnt[j - 1] + 1;
        chkmax(maxl, f[i][j]);
      }
    }
    for (int j = i; j <= n; ++j) {
      if (a[j] == 2)
        cntf[i][j] = max(f[i][j], cntf[i][j - 1]);
      else
        cntf[i][j] = cntf[i][j - 1];
    }
  }
  for (int i = 1; i <= n; ++i) chkmax(ans, f[1][i]);
}
inline void build_g() {
  for (int i = 1; i <= n - 1; ++i) {
    g[i][i] = 1;
    memset(cnt, 0, sizeof(cnt));
    if (a[i] == 1) cnt[i] = 1;
    int maxl = 1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 2) {
        g[i][j] = max(maxl, cnt[j - 1]) + 1;
        cnt[j] = cnt[j - 1];
        chkmax(maxl, g[i][j]);
      } else {
        g[i][j] = cnt[j - 1] + 1;
        cnt[j] = cnt[j - 1] + 1;
        chkmax(maxl, g[i][j]);
      }
    }
    for (int j = i; j <= n; ++j) {
      if (a[j] == 2)
        cntg[i][j] = max(g[i][j], cntg[i][j - 1]);
      else
        cntg[i][j] = cntg[i][j - 1];
    }
  }
}
inline void solve() {
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      int p = max(cntg[n - j + 1][n - i + 1] + s1[i - 1],
                  max(s1[i - 1] + s2[j] - s2[i - 1],
                      cntf[1][i - 1] + s2[j] - s2[i - 1]));
      chkmax(ans, max(cntf[j + 1][n] + s1[j],
                      max(s1[j] + s2[n] - s2[j], p + s2[n] - s2[j])));
    }
  chkmax(ans, s1[n]);
  chkmax(ans, s2[n]);
  printf("%d\n", ans);
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(a[i]);
  build_s();
  build_f();
  int i = 1, j = n;
  while (i < j) {
    swap(a[i], a[j]);
    i++;
    j--;
  }
  build_g();
  solve();
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
template <typename T>
bool chkmax(T& a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
bool chkmin(T& a, T b) {
  return a > b ? a = b, 1 : 0;
}
const int maxn = 2000 + 10;
int n, ans;
int a[maxn], cnt[maxn];
int s1[maxn], s2[maxn];
int f[maxn][maxn], cntf[maxn][maxn];
int g[maxn][maxn], cntg[maxn][maxn];
inline void read(int& x) {
  x = 0;
  int p = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - '0');
    c = getchar();
  }
  x *= p;
}
inline void file() {
  freopen("C.in", "r", stdin);
  freopen("C.out", "w", stdout);
}
inline void build_s() {
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      s1[i] = s1[i - 1] + 1;
      s2[i] = s2[i - 1];
    } else {
      s1[i] = s1[i - 1];
      s2[i] = s2[i - 1] + 1;
    }
  }
}
inline void build_f() {
  for (int i = 1; i <= n - 1; ++i) {
    f[i][i] = 1;
    memset(cnt, 0, sizeof(cnt));
    if (a[i] == 1) cnt[i] = 1;
    int maxl = 1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 2) {
        f[i][j] = max(maxl, cnt[j - 1]) + 1;
        cnt[j] = cnt[j - 1];
        chkmax(maxl, f[i][j]);
      } else {
        f[i][j] = cnt[j - 1] + 1;
        cnt[j] = cnt[j - 1] + 1;
        chkmax(maxl, f[i][j]);
      }
    }
    for (int j = i; j <= n; ++j) {
      if (a[j] == 2)
        cntf[i][j] = max(f[i][j], cntf[i][j - 1]);
      else
        cntf[i][j] = cntf[i][j - 1];
    }
  }
  for (int i = 1; i <= n; ++i) chkmax(ans, f[1][i]);
}
inline void build_g() {
  for (int i = 1; i <= n - 1; ++i) {
    g[i][i] = 1;
    memset(cnt, 0, sizeof(cnt));
    if (a[i] == 1) cnt[i] = 1;
    int maxl = 1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 2) {
        g[i][j] = max(maxl, cnt[j - 1]) + 1;
        cnt[j] = cnt[j - 1];
        chkmax(maxl, g[i][j]);
      } else {
        g[i][j] = cnt[j - 1] + 1;
        cnt[j] = cnt[j - 1] + 1;
        chkmax(maxl, g[i][j]);
      }
    }
    for (int j = i; j <= n; ++j) {
      if (a[j] == 2)
        cntg[i][j] = max(g[i][j], cntg[i][j - 1]);
      else
        cntg[i][j] = cntg[i][j - 1];
    }
  }
}
inline void solve() {
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      int p = max(cntg[n - j + 1][n - i + 1] + s1[i - 1],
                  max(s1[i - 1] + s2[j] - s2[i - 1],
                      cntf[1][i - 1] + s2[j] - s2[i - 1]));
      chkmax(ans, max(cntf[j + 1][n] + s1[j],
                      max(s1[j] + s2[n] - s2[j], p + s2[n] - s2[j])));
    }
  chkmax(ans, s1[n]);
  chkmax(ans, s2[n]);
  printf("%d\n", ans);
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(a[i]);
  build_s();
  build_f();
  int i = 1, j = n;
  while (i < j) {
    swap(a[i], a[j]);
    i++;
    j--;
  }
  build_g();
  solve();
  return 0;
}

```