#include <bits/stdc++.h>
using namespace std;
int setBit(int n, int pos) { return n = n | (1 << pos); }
int resetBit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkBit(long long n, long long pos) { return (bool)(n & (1LL << pos)); }
const int MAX = 2010;
int ara[MAX];
int dp[MAX][MAX][4];
int call(int pos, int n, int last) {
  if (pos < n) return 0;
  if (dp[pos][n][last] != -1) return dp[pos][n][last];
  int ret = call(pos - 1, n, last);
  if (ara[pos] >= last) {
    ret = max(ret, 1 + call(pos - 1, n, ara[pos]));
  }
  return dp[pos][n][last] = ret;
}
int cum1[MAX], cum2[MAX];
int main() {
  int n, cnt1 = 0, cnt2 = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ara[i]);
  }
  for (int i = 1; i <= n; i++) {
    cum1[i] = cum1[i - 1];
    cum2[i] = cum2[i - 1];
    if (ara[i] == 1) {
      cum1[i]++;
      cnt1++;
    } else {
      cum2[i]++;
      cnt2++;
    }
  }
  memset(dp, -1, sizeof(dp));
  reverse(ara + 1, ara + n + 1);
  int mx = call(n, 1, 1), lala;
  memset(dp, -1, sizeof(dp));
  reverse(ara + 1, ara + n + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      lala = cum1[i - 1];
      lala += call(j, i, 1);
      lala += cnt2 - cum2[j];
      mx = max(mx, lala);
    }
  }
  printf("%d\n", mx);
  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>
using namespace std;
int setBit(int n, int pos) { return n = n | (1 << pos); }
int resetBit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkBit(long long n, long long pos) { return (bool)(n & (1LL << pos)); }
const int MAX = 2010;
int ara[MAX];
int dp[MAX][MAX][4];
int call(int pos, int n, int last) {
  if (pos < n) return 0;
  if (dp[pos][n][last] != -1) return dp[pos][n][last];
  int ret = call(pos - 1, n, last);
  if (ara[pos] >= last) {
    ret = max(ret, 1 + call(pos - 1, n, ara[pos]));
  }
  return dp[pos][n][last] = ret;
}
int cum1[MAX], cum2[MAX];
int main() {
  int n, cnt1 = 0, cnt2 = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ara[i]);
  }
  for (int i = 1; i <= n; i++) {
    cum1[i] = cum1[i - 1];
    cum2[i] = cum2[i - 1];
    if (ara[i] == 1) {
      cum1[i]++;
      cnt1++;
    } else {
      cum2[i]++;
      cnt2++;
    }
  }
  memset(dp, -1, sizeof(dp));
  reverse(ara + 1, ara + n + 1);
  int mx = call(n, 1, 1), lala;
  memset(dp, -1, sizeof(dp));
  reverse(ara + 1, ara + n + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      lala = cum1[i - 1];
      lala += call(j, i, 1);
      lala += cnt2 - cum2[j];
      mx = max(mx, lala);
    }
  }
  printf("%d\n", mx);
  return 0;
}

```