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;

template<class T> void vin(vector<T>& v, int n) {
    v.resize(n);
    for (int i = 0; i < n; ++i) {
        cin >> v[i];
    }
}

double m[50000];

int main() {
    int N;
    cin >> N;
    vector<double> t, v;
    vin(t, N);
    vin(v, N);
    rep(i, 50000) {
        m[i] = 1e8;
    }
    int sum = 0;
    rep(i, N) {
        t[i] *= 2;
        for (int j = sum; j <= sum + t[i]; ++j) {
            m[j] = min(m[j], 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 S = 0;
    rep(i, sum) {
        S += m[i] + m[i + 1];
    }
    printf("%.12lf\n", (S / 2.0) / 2.0);
    return 0;
}