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;

double t[105], v[105], st[105];
int main(){
    int n;
    cin >> n;
    double sum = 0.0;
    for(int i = 0; i < n; i++){
        cin >> t[i+1];
        st[i+1] = st[i] + t[i+1];
    }
    st[n+1] += st[n];
    for(int i = 0; i < n; i++) cin >> v[i+1];
    double ans = 0.0, prev = 0.0;
    for(double i = 0.5; i <= st[n]; i += 0.5){
        double cur = 100;
        for(int j = 0; j <= n+1; j++){
            if(j >= 1 && st[j-1] <= i && i <= st[j]){
                cur = min(cur, (double)v[j]);
            }
            if(st[j] < i){
                cur = min(cur, (double)(v[j] + (i - st[j])));
            }
            if(j >= 1 && st[j-1] > i){
                cur = min(cur, (double)(v[j] + (st[j-1] - i)));
            }
        }
        ans += (cur + prev) * 0.25;
        prev = cur;
    }
    cout << fixed << setprecision(9) << ans << endl;
    return 0;
}