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;


void solve() {
    int n; cin >> n;
    vector<double> t(n), v(n);
    for (auto& x: t) {
        cin >> x;
    }
    for (auto& x: v) {
        cin >> x;
    }
    t.emplace_back(0);
    v.emplace_back(0);
    double res = 0;
    double cur = 0;
    for (int i = 0; i < n; i++) {
        double out = v[i];
        double dt = 0;
        for (int j = i+1; j <= n; j++) {
            out = min(out, v[j] + dt);
            dt += t[j];
        }
        double inc = min({v[i]-cur, t[i], (out+t[i]-cur)/2.0});
        double dec = max(0.0, cur+inc - out);
        double cons = t[i] - inc - dec;
        res += (cur*2.+inc)*inc/2.;
        cur += inc;
        res += cur * cons;
        res += (cur*2.-dec)*dec/2.;
        cur -= dec;
    }
    cout << fixed << setprecision(5) << res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}