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;

using ll = long long;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<int> t(n + 1, 0), v(n);
    for (int i = 0; i < n; i++) {
        cin >> t[i + 1];
        t[i + 1] += t[i];
    }
    for (int i = 0; i < n; i++) cin >> v[i];
    double ans = 0, cur = 0;
    int tsum2 = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < (t[i + 1] - t[i]) * 2; j++) {
            cur = min({cur + 0.5, (double)v[i], t[n] - t[i] - (j + 1) * 0.5});
            for (int k = i + 1; k < n; k++) {
                cur = min(cur, t[k] - t[i] - (j + 1) * 0.5 + v[k]);
            }
            ans += cur;
        }
    }
    ans /= 2;
    cout << fixed << setprecision(10) << ans << endl;
    return 0;
}