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)<(int)(n);++(i))
using ll = long long;
using P = pair< ll, ll >;

using namespace std;

int N;
int T[202];
double V[202];
double M[50000];

int main() {
    cin >> N;
    int sum = 0;
    rep(i, 50000) M[i] = 2020;
    rep(i, N) cin >> T[i];
    rep(i, N) cin >> V[i];
    rep(i, N) {
        T[i] *= 2;
        for (int x=sum; x<=sum+T[i]; ++x) M[x] = min(M[x], V[i]);
        sum += T[i];
    }
    M[0] = M[sum] = 0;
    rep(i, sum) M[i+1] = min(M[i+1], M[i]+0.5);
    for (int i=sum-1; i>=0; --i) M[i] = min(M[i], M[i+1] + 0.5);
    double R = 0;
    rep(i, sum) R += M[i] + M[i+1];
    printf("%.12lf\n", R/4.0);
}