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 <stdio.h>
#define MAX_N 100 + 2
#define MAX_T 40000

int t[MAX_N];
double v[MAX_N], f[MAX_T];

double min(double a, double b)
{
    return a < b? a: b;
}

int main(void)
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d", &t[i]);
    for (int i = 0; i < n; i++)
        scanf("%lf", &v[i]);
    for (int i = 0; i < MAX_T; i++)
        f[i] = 1.E8;
    int tot = 0;
    for (int i = 0; i < n; i++) {
        t[i] *= 2;
        for (int j = tot; j <= tot + t[i]; j++)
            f[j] = min(f[j], v[i]);
        tot += t[i];
    }
    f[0] = f[tot] = 0.;
    for (int i = 1; i <= tot; i++)
        f[i] = min(f[i], f[i - 1] + .5);
    for (int i = tot - 1; i >= 0; i--)
        f[i] = min(f[i], f[i + 1] + .5);
    double ans = 0.;
    for (int i = 0; i < tot; i++)
        ans += .25 * (f[i] + f[i + 1]);
    printf("%f\n", ans);
    return 0;
}