In the year 2168, AtCoder Inc., which is much larger than now, is starting a limited express train service called AtCoder Express.

In the plan developed by the president Takahashi, the trains will run as follows:

* A train will run for (t_1 + t_2 + t_3 + ... + t_N) seconds.
* In the first t_1 seconds, a train must run at a speed of at most v_1 m/s (meters per second). Similarly, in the subsequent t_2 seconds, a train must run at a speed of at most v_2 m/s, and so on.



According to the specifications of the trains, the acceleration of a train must be always within ±1m/s^2. Additionally, a train must stop at the beginning and the end of the run.

Find the maximum possible distance that a train can cover in the run.

Constraints

* 1 \leq N \leq 100
* 1 \leq t_i \leq 200
* 1 \leq v_i \leq 100
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N
t_1 t_2 t_3 … t_N
v_1 v_2 v_3 … v_N


Output

Print the maximum possible that a train can cover in the run.
Output is considered correct if its absolute difference from the judge's output is at most 10^{-3}.

Examples

Input

1
100
30


Output

2100.000000000000000


Input

2
60 50
34 38


Output

2632.000000000000000


Input

3
12 14 2
6 2 7


Output

76.000000000000000


Input

1
9
10


Output

20.250000000000000000


Input

10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70


Output

20291.000000000000
#include <bits/stdc++.h>
using namespace std;

const int inf = 1000000007;

int T[102], V[20004], dp[20004][102];

int solve(int x, int y)
{
    if(y < 0 || y > V[x])
        return -inf;

    if(x == 0)
        return 0;

    if(dp[x][y] != -1)
        return dp[x][y];
    
    int ans = solve(x-1, y) + y * 4;
    ans = max(ans, solve(x-1, y-1) + (y-1) * 4 + 2);
    
    if(y != V[x])
    {
        ans = max(ans, solve(x-1, y+1) + y * 4 + 2);
        ans = max(ans, solve(x-1, y) + y * 4 + 1);
    }

    return dp[x][y] = ans;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(NULL), cout.tie(NULL);

    int N;
    cin >> N;

    for(int i=1; i<=N; i++)
        cin >> T[i];

    int k = 0;
    
    for(int i=1; i<=N; i++)
    {
        int v;
        cin >> v;

        while(T[i]--)
            V[++k] = v;
    }

    memset(dp, -1, sizeof(dp));

    cout << fixed << setprecision(6) << solve(k, 0) / 4.0 << "\n";
    return 0;
}
