#include <bits/stdc++.h>
using namespace std;
const long long oo = 1ll << 17;
const long long mos = 0x7FFFFFFF;
const long long nmos = 0x80000000;
const int inf = 0x3f3f3f3f;
const long long inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399;
const double tPHI = 0.38196601;
template <typename T>
inline T read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
inline void cmax(int &x, int y) {
  if (x < y) x = y;
}
inline void cmax(long long &x, long long y) {
  if (x < y) x = y;
}
inline void cmin(int &x, int y) {
  if (x > y) x = y;
}
inline void cmin(long long &x, long long y) {
  if (x > y) x = y;
}
const int maxn = 2009;
int a[maxn], cnt1[maxn], cnt2[maxn];
int t[maxn << 2];
void build(int l, int r, int rt) {
  if (l == r) {
    t[rt] = cnt1[l] + cnt2[l + 1];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  t[rt] = max(t[rt << 1], t[rt << 1 | 1]);
}
int query(int L, int R, int l, int r, int rt) {
  if (l >= L && r <= R) {
    return t[rt];
  }
  int mx = 0;
  int mid = (l + r) >> 1;
  if (mid >= L) mx = max(mx, query(L, R, l, mid, rt << 1));
  if (mid < R) mx = max(mx, query(L, R, mid + 1, r, rt << 1 | 1));
  return mx;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%d", &a[i]);
  for (int i = (1); i <= (n); ++i) {
    if (a[i] == 1)
      cnt1[i] = cnt1[i - 1] + 1;
    else
      cnt1[i] = cnt1[i - 1];
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] == 2)
      cnt2[i] = cnt2[i + 1] + 1;
    else
      cnt2[i] = cnt2[i + 1];
  }
  build(0, n, 1);
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    int tmp =
        query(0, i, 0, n, 1) + query(i, n, 0, n, 1) - cnt1[i] - cnt2[i + 1];
    ans = max(ans, tmp);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 long long oo = 1ll << 17;
const long long mos = 0x7FFFFFFF;
const long long nmos = 0x80000000;
const int inf = 0x3f3f3f3f;
const long long inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399;
const double tPHI = 0.38196601;
template <typename T>
inline T read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
inline void cmax(int &x, int y) {
  if (x < y) x = y;
}
inline void cmax(long long &x, long long y) {
  if (x < y) x = y;
}
inline void cmin(int &x, int y) {
  if (x > y) x = y;
}
inline void cmin(long long &x, long long y) {
  if (x > y) x = y;
}
const int maxn = 2009;
int a[maxn], cnt1[maxn], cnt2[maxn];
int t[maxn << 2];
void build(int l, int r, int rt) {
  if (l == r) {
    t[rt] = cnt1[l] + cnt2[l + 1];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  t[rt] = max(t[rt << 1], t[rt << 1 | 1]);
}
int query(int L, int R, int l, int r, int rt) {
  if (l >= L && r <= R) {
    return t[rt];
  }
  int mx = 0;
  int mid = (l + r) >> 1;
  if (mid >= L) mx = max(mx, query(L, R, l, mid, rt << 1));
  if (mid < R) mx = max(mx, query(L, R, mid + 1, r, rt << 1 | 1));
  return mx;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%d", &a[i]);
  for (int i = (1); i <= (n); ++i) {
    if (a[i] == 1)
      cnt1[i] = cnt1[i - 1] + 1;
    else
      cnt1[i] = cnt1[i - 1];
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] == 2)
      cnt2[i] = cnt2[i + 1] + 1;
    else
      cnt2[i] = cnt2[i + 1];
  }
  build(0, n, 1);
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    int tmp =
        query(0, i, 0, n, 1) + query(i, n, 0, n, 1) - cnt1[i] - cnt2[i + 1];
    ans = max(ans, tmp);
  }
  cout << ans << endl;
  return 0;
}

```