#include <bits/stdc++.h>
int input[2001];
int number1[2001], number2[2001];
int lnds12[2001][2001], lnis21[2001][2001];
int n;
int inline max(int a, int b) { return (a > b) ? a : b; }
int lnds11(int i, int j) {
  if (j < i) return 0;
  return number1[j] - number1[i - 1];
}
int lnds22(int i, int j) {
  if (j < i) return 0;
  return number2[j] - number2[i - 1];
}
int dp_lnds12(int i, int j) {
  if (j < i) return 0;
  if (lnds12[i][j] != -1) return lnds12[i][j];
  if (j - i < 1) return lnds12[i][j] = 0;
  if (j - i == 1) {
    if (input[j] == 2 && input[i] == 1) return lnds12[i][j] = 2;
    return lnds12[i][j] = 0;
  }
  if (input[j] == 2)
    return lnds12[i][j] = max(dp_lnds12(i, j - 1) + 1, lnds11(i, j - 1) + 1);
  return lnds12[i][j] = dp_lnds12(i, j - 1);
}
int dp_lnis21(int i, int j) {
  if (j < i) return 0;
  if (lnis21[i][j] != -1) return lnis21[i][j];
  if (j - i < 1) return lnis21[i][j] = 0;
  if (j - i == 1) {
    if (input[j] == 1 && input[i] == 2) return lnis21[i][j] = 2;
    return lnis21[i][j] = 0;
  }
  if (input[j] == 1)
    return lnis21[i][j] = max(dp_lnis21(i, j - 1) + 1, lnds22(i, j - 1) + 1);
  return lnis21[i][j] = dp_lnis21(i, j - 1);
}
int main() {
  memset(lnds12, -1, sizeof(lnds12));
  memset(lnis21, -1, sizeof(lnis21));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &input[i]);
    if (input[i] == 1) {
      number1[i] = number1[i - 1] + 1;
      number2[i] = number2[i - 1];
    } else {
      number1[i] = number1[i - 1];
      number2[i] = number2[i - 1] + 1;
    }
  }
  int ans = 0;
  for (int l = 1; l <= n; l++) {
    for (int r = l; r <= n; r++) {
      ans = max(ans, lnds11(1, l - 1) + lnds11(l, r) + lnds11(r + 1, n));
      ans = max(ans, lnds11(1, l - 1) + lnds11(l, r) + dp_lnds12(r + 1, n));
      ans = max(ans, lnds11(1, l - 1) + lnds11(l, r) + lnds22(r + 1, n));
      ans = max(ans, lnds11(1, l - 1) + dp_lnis21(l, r) + lnds22(r + 1, n));
      ans = max(ans, lnds11(1, l - 1) + lnds22(l, r) + lnds22(r + 1, n));
      ans = max(ans, dp_lnds12(1, l - 1) + lnds22(l, r) + lnds22(r + 1, n));
      ans = max(ans, lnds22(1, l - 1) + lnds22(l, r) + lnds22(r + 1, n));
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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>
int input[2001];
int number1[2001], number2[2001];
int lnds12[2001][2001], lnis21[2001][2001];
int n;
int inline max(int a, int b) { return (a > b) ? a : b; }
int lnds11(int i, int j) {
  if (j < i) return 0;
  return number1[j] - number1[i - 1];
}
int lnds22(int i, int j) {
  if (j < i) return 0;
  return number2[j] - number2[i - 1];
}
int dp_lnds12(int i, int j) {
  if (j < i) return 0;
  if (lnds12[i][j] != -1) return lnds12[i][j];
  if (j - i < 1) return lnds12[i][j] = 0;
  if (j - i == 1) {
    if (input[j] == 2 && input[i] == 1) return lnds12[i][j] = 2;
    return lnds12[i][j] = 0;
  }
  if (input[j] == 2)
    return lnds12[i][j] = max(dp_lnds12(i, j - 1) + 1, lnds11(i, j - 1) + 1);
  return lnds12[i][j] = dp_lnds12(i, j - 1);
}
int dp_lnis21(int i, int j) {
  if (j < i) return 0;
  if (lnis21[i][j] != -1) return lnis21[i][j];
  if (j - i < 1) return lnis21[i][j] = 0;
  if (j - i == 1) {
    if (input[j] == 1 && input[i] == 2) return lnis21[i][j] = 2;
    return lnis21[i][j] = 0;
  }
  if (input[j] == 1)
    return lnis21[i][j] = max(dp_lnis21(i, j - 1) + 1, lnds22(i, j - 1) + 1);
  return lnis21[i][j] = dp_lnis21(i, j - 1);
}
int main() {
  memset(lnds12, -1, sizeof(lnds12));
  memset(lnis21, -1, sizeof(lnis21));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &input[i]);
    if (input[i] == 1) {
      number1[i] = number1[i - 1] + 1;
      number2[i] = number2[i - 1];
    } else {
      number1[i] = number1[i - 1];
      number2[i] = number2[i - 1] + 1;
    }
  }
  int ans = 0;
  for (int l = 1; l <= n; l++) {
    for (int r = l; r <= n; r++) {
      ans = max(ans, lnds11(1, l - 1) + lnds11(l, r) + lnds11(r + 1, n));
      ans = max(ans, lnds11(1, l - 1) + lnds11(l, r) + dp_lnds12(r + 1, n));
      ans = max(ans, lnds11(1, l - 1) + lnds11(l, r) + lnds22(r + 1, n));
      ans = max(ans, lnds11(1, l - 1) + dp_lnis21(l, r) + lnds22(r + 1, n));
      ans = max(ans, lnds11(1, l - 1) + lnds22(l, r) + lnds22(r + 1, n));
      ans = max(ans, dp_lnds12(1, l - 1) + lnds22(l, r) + lnds22(r + 1, n));
      ans = max(ans, lnds22(1, l - 1) + lnds22(l, r) + lnds22(r + 1, n));
    }
  }
  printf("%d\n", ans);
  return 0;
}

```