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;
#define rep(i,n) for(int i=0;i<n;++i)
#define rep2(i,i0,n) for(int i=i0;i<n;++i)

int main()
{
    int n;
    cin >> n;
    vector<double> t(n), v(n);
    for(auto& ti:t) cin >> ti;
    for(auto& vi:v) cin >> vi;

    vector<double> y0(n+1,0);
    double st = 0;
    rep(i,n) {
        y0[i] = v[i] + st;
        st += t[i];
    }
    y0[n] = st;

    double d = 0;
    double cv = 0;
    double ct = 0;
    double dt = 0.5;

    rep(i,n) {
        double min_y0 = 1e9;
        rep2(j,i+1,n+1) 
            min_y0 = min(min_y0, y0[j]);

        rep(j,t[i]*2) {            
            ct += dt;
            double limit = min(v[i], min_y0-ct);
            double pv = cv;
            cv = min(cv+dt, limit);
            d += (pv + cv) * dt / 2;
        }
    }

    cout << fixed << setprecision(12) << d << endl;
    return 0;
}