#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2010;
int dp[MAXN][MAXN], n, a[MAXN], num2[MAXN], num1[MAXN], num3[MAXN];
int dfs(int l, int r) {
  if (dp[l][r] != -1) {
    return dp[l][r];
  }
  if (l == r) {
    dp[l][r] = 1;
    return dp[l][r];
  }
  if (a[l] == 2) {
    dp[l][r] = dfs(l + 1, r) + 1;
  } else {
    dp[l][r] = max(num3[l] - num3[r + 1], dfs(l + 1, r));
  }
  return dp[l][r];
}
void ceshi() {
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      printf("%d ", dp[i][j]);
    }
    printf("\n");
  }
}
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(dp, -1, sizeof(dp));
    memset(num2, 0, sizeof(num2));
    memset(num1, 0, sizeof(num1));
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
    }
    for (int i = n; i >= 1; --i) {
      if (a[i] == 2) {
        num2[i] = num2[i + 1] + 1;
      } else {
        num2[i] = num2[i + 1];
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 1) {
        num1[i] = num1[i - 1] + 1;
      } else {
        num1[i] = num1[i - 1];
      }
    }
    memset(num3, 0, sizeof(num3));
    for (int i = n; i >= 1; --i) {
      if (a[i] == 1) {
        num3[i] = num3[i + 1] + 1;
      } else {
        num3[i] = num3[i + 1];
      }
    }
    memset(dp, -1, sizeof(dp));
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = i; j <= n; ++j) {
        if (dp[i][j] == -1) {
          dp[i][j] = dfs(i, j);
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = i; j <= n; ++j) {
        ans = max(ans, num1[i - 1] + dp[i][j] + num2[j + 1]);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create 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 MAXN = 2010;
int dp[MAXN][MAXN], n, a[MAXN], num2[MAXN], num1[MAXN], num3[MAXN];
int dfs(int l, int r) {
  if (dp[l][r] != -1) {
    return dp[l][r];
  }
  if (l == r) {
    dp[l][r] = 1;
    return dp[l][r];
  }
  if (a[l] == 2) {
    dp[l][r] = dfs(l + 1, r) + 1;
  } else {
    dp[l][r] = max(num3[l] - num3[r + 1], dfs(l + 1, r));
  }
  return dp[l][r];
}
void ceshi() {
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      printf("%d ", dp[i][j]);
    }
    printf("\n");
  }
}
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(dp, -1, sizeof(dp));
    memset(num2, 0, sizeof(num2));
    memset(num1, 0, sizeof(num1));
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
    }
    for (int i = n; i >= 1; --i) {
      if (a[i] == 2) {
        num2[i] = num2[i + 1] + 1;
      } else {
        num2[i] = num2[i + 1];
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 1) {
        num1[i] = num1[i - 1] + 1;
      } else {
        num1[i] = num1[i - 1];
      }
    }
    memset(num3, 0, sizeof(num3));
    for (int i = n; i >= 1; --i) {
      if (a[i] == 1) {
        num3[i] = num3[i + 1] + 1;
      } else {
        num3[i] = num3[i + 1];
      }
    }
    memset(dp, -1, sizeof(dp));
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = i; j <= n; ++j) {
        if (dp[i][j] == -1) {
          dp[i][j] = dfs(i, j);
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = i; j <= n; ++j) {
        ans = max(ans, num1[i - 1] + dp[i][j] + num2[j + 1]);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

```