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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include<iomanip>
typedef long long ll;
using namespace std;

double t[105], v[105], sum[105];

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> t[i+1];
        sum[i+1] = sum[i] + t[i+1];
    }
    for (int i = 0; i < n; i++) {
        cin >> v[i+1];
    }

    double ans=0, p=0;
    int f = 0;

    for (double i = 0.5; i <= sum[n]; i+=0.5) {
        if (i > sum[f]) f++;
        double now = v[f];
        for (int j = 0; j <= n+1; j++) {
            double c = now;
            if (j < f) {
                c = v[j] + i - sum[j];
            } else if (j > f) {
                c = v[j] + sum[j-1] - i;
            }
            now = min(now, c);
        }
        ans += (p+now)/4.0;
        p = now;
    }

    cout << fixed << setprecision(9) << ans << endl;
    return 0;
}