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;
#define rep(i, n) for (int i=0; i<n; i++)
#define pb push_back
#define int long long

int N;
long double t[110], v[110], lim[110];

signed main() {
    cin.tie(0); ios::sync_with_stdio(false);
    cin >> N;
    rep(i, N) cin >> t[i];
    rep(i, N) cin >> v[i];
    for (int i=N-2; i>=0; i--) lim[i] = min({lim[i+1]+t[i+1], v[i], v[i+1]});
    double ans = 0, now_spe = 0;
    rep(i, N) {
        if (now_spe+t[i]<lim[i]) {
            ans += (2*now_spe+t[i])*t[i]/2.0;
            now_spe += t[i];
        }
        else {
            double max_spe = min(v[i], (t[i]+lim[i]+now_spe)/2);
            ans += (now_spe+max_spe)*(max_spe-now_spe)/2;
            ans += (lim[i]+max_spe)*(max_spe-lim[i])/2;
            ans += max_spe*(t[i]-(max_spe-now_spe)-(max_spe-lim[i]));
            now_spe = lim[i];
        }
    }
    cout << fixed << setprecision(10) << ans << endl;
}