#include <bits/stdc++.h>
using namespace std;
const int N = 2001;
const int MOD = 1e9 + 7;
int n, a[N], o[N], t[N], dp1[N][N], dp2[N][N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == 1)
      o[i] = 1;
    else
      t[i] = 1;
    if (i) {
      o[i] += o[i - 1];
      t[i] += t[i - 1];
    }
  }
  int lastOne, lastTwo;
  for (int j = 0; j < n; j++) {
    dp1[j][j] = 1;
    lastOne = (a[j] == 1 ? j : -1);
    lastTwo = (a[j] == 2 ? j : -1);
    for (int i = j + 1; i < n; i++) {
      if (a[i] == 1) {
        dp1[j][i] = (lastOne == -1 ? 1 : dp1[j][lastOne] + 1);
        lastOne = i;
      } else {
        dp1[j][i] = (lastTwo == -1 ? 1 : dp1[j][lastTwo] + 1);
        dp1[j][i] = max(dp1[j][i], (lastOne == -1 ? 1 : dp1[j][lastOne] + 1));
        lastTwo = i;
      }
    }
  }
  for (int j = 0; j < n; j++) {
    dp2[j][j] = 1;
    lastOne = (a[j] == 1 ? j : -1);
    lastTwo = (a[j] == 2 ? j : -1);
    for (int i = j + 1; i < n; i++) {
      if (a[i] == 1) {
        dp2[j][i] = (lastOne == -1 ? 1 : dp2[j][lastOne] + 1);
        dp2[j][i] = max(dp2[j][i], (lastTwo == -1 ? 1 : dp2[j][lastTwo] + 1));
        lastOne = i;
      } else {
        dp2[j][i] = (lastTwo == -1 ? 1 : dp2[j][lastTwo] + 1);
        lastTwo = i;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      int A = 0, B = 0, C = 0;
      if (i) A = o[i - 1];
      if (j + 1 < n) C = t[n - 1] - (j + 1 > 0 ? t[j] : 0);
      B = max(dp2[i][j],
              max(o[j] - (i ? o[i - 1] : 0), t[j] - (i ? t[i - 1] : 0)));
      ans = max(ans, A + B + C);
      A = 0, B = 0, C = 0;
      if (i) A = max(t[i - 1], dp1[0][i - 1]);
      if (j + 1 < n) C = t[n - 1] - (j + 1 > 0 ? t[j] : 0);
      B = t[j] - (i ? t[i - 1] : 0);
      ans = max(ans, A + B + C);
      A = 0, B = 0, C = 0;
      if (i) A = o[i - 1];
      if (j + 1 < n) C = max(dp1[j + 1][n - 1], t[n - 1] - (j + 1 ? t[j] : 0));
      B = o[j] - (i ? o[i - 1] : 0);
      ans = max(ans, A + B + C);
    }
  }
  printf("%d\n", ans);
  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;
const int N = 2001;
const int MOD = 1e9 + 7;
int n, a[N], o[N], t[N], dp1[N][N], dp2[N][N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == 1)
      o[i] = 1;
    else
      t[i] = 1;
    if (i) {
      o[i] += o[i - 1];
      t[i] += t[i - 1];
    }
  }
  int lastOne, lastTwo;
  for (int j = 0; j < n; j++) {
    dp1[j][j] = 1;
    lastOne = (a[j] == 1 ? j : -1);
    lastTwo = (a[j] == 2 ? j : -1);
    for (int i = j + 1; i < n; i++) {
      if (a[i] == 1) {
        dp1[j][i] = (lastOne == -1 ? 1 : dp1[j][lastOne] + 1);
        lastOne = i;
      } else {
        dp1[j][i] = (lastTwo == -1 ? 1 : dp1[j][lastTwo] + 1);
        dp1[j][i] = max(dp1[j][i], (lastOne == -1 ? 1 : dp1[j][lastOne] + 1));
        lastTwo = i;
      }
    }
  }
  for (int j = 0; j < n; j++) {
    dp2[j][j] = 1;
    lastOne = (a[j] == 1 ? j : -1);
    lastTwo = (a[j] == 2 ? j : -1);
    for (int i = j + 1; i < n; i++) {
      if (a[i] == 1) {
        dp2[j][i] = (lastOne == -1 ? 1 : dp2[j][lastOne] + 1);
        dp2[j][i] = max(dp2[j][i], (lastTwo == -1 ? 1 : dp2[j][lastTwo] + 1));
        lastOne = i;
      } else {
        dp2[j][i] = (lastTwo == -1 ? 1 : dp2[j][lastTwo] + 1);
        lastTwo = i;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      int A = 0, B = 0, C = 0;
      if (i) A = o[i - 1];
      if (j + 1 < n) C = t[n - 1] - (j + 1 > 0 ? t[j] : 0);
      B = max(dp2[i][j],
              max(o[j] - (i ? o[i - 1] : 0), t[j] - (i ? t[i - 1] : 0)));
      ans = max(ans, A + B + C);
      A = 0, B = 0, C = 0;
      if (i) A = max(t[i - 1], dp1[0][i - 1]);
      if (j + 1 < n) C = t[n - 1] - (j + 1 > 0 ? t[j] : 0);
      B = t[j] - (i ? t[i - 1] : 0);
      ans = max(ans, A + B + C);
      A = 0, B = 0, C = 0;
      if (i) A = o[i - 1];
      if (j + 1 < n) C = max(dp1[j + 1][n - 1], t[n - 1] - (j + 1 ? t[j] : 0));
      B = o[j] - (i ? o[i - 1] : 0);
      ans = max(ans, A + B + C);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```