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>
using namespace std;

int veloc(int l, int r, int v, int x)
{
    if (x <= l) return v + (l - x);
    else if (x >= r) return v + (x - r);
    return v;
}

int main()
{
    int n, t;
    cin >> n;

    int l[n + 2], r[n + 2], v[n + 2];

    l[0] = r[0] = l[1] = 0;
    for (int i = 1; i <= n; i++) {
        cin >> t;
        l[i] = r[i - 1];
        r[i] = l[i] + t * 2;
    }
    l[n + 1] = r[n + 1] = r[n];

    v[0] = 0;
    for (int i = 1; i <= n; i++) { cin >> v[i]; v[i] *= 2; }
    v[n + 1] = 0;

    int pr = 0;
    int ans = 0;
    for (int x = 1; x <= r[n]; x++) {
        int mn = 1e9;
        for (int i = 0; i <= n + 1; i++) {
            mn = min(mn, veloc(l[i], r[i], v[i], x));
        }
        ans += pr + mn;
        pr = mn;
    }
    printf("%lf\n", double(ans) / 8);
}