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;

int main(void) {
    cout << fixed << setprecision(16);

    ll n;
    cin >> n;
    vector<double> t(n+2), v(n+2);
    for(ll i=0; i<n; i++) cin >> t[i+1];
    for(ll i=0; i<n; i++) cin >> v[i+1];

    ll ts = accumulate(t.begin(), t.end(), 0ll);
    partial_sum(t.begin(), t.end(), t.begin());

    double s = 0, p = 0;
    for(ll i=0; i<=2*ts; i++) {
        double tc = 0.5*i;
        double k = p + 0.5;
        {
            ll j=0;
            while(t[j] < tc)
                k = min(k, (tc - t[j]) + v[j]), j++;
            k = min(k, v[j]);
            while(j < n+1)
                k = min(k, (t[j] - tc) + v[j+1]), j++;
        }
        s += (p + k) * 0.5 / 2;
        p = k;
    }
    cout << s << endl;
}