Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ld long double
#define IO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
using namespace std;
const int N = 105;
int n, p;
ld t[N], v[N], s[N], ans, curv;
int main()
{
    IO;
    cin >> n;
    for(int i = 1; i <= n; i++)
    {
        cin >> t[i];
        s[i] = s[i - 1] + t[i];
    }
    for(int i = 1; i <= n; i++)
        cin >> v[i];
    for(ld i = 0.5; i <= s[n]; i += 0.5)
    {
        while(i > s[p])
            p++;
        ld vel = v[p];
        for(int j = 0; j < n + 2; j++)
        {
            if(j < p)
                vel = fminl(vel, v[j] + i - s[j]);
            else if(j > p)
                vel = fminl(vel, v[j] + s[j - 1] - i);
        }
        ans += (vel + curv) / 2;
        curv = vel;
    }
    cout << fixed << setprecision(15) << ans / 2;
}