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 INF 1001001001

int main() {
    int N;
    cin >> N;
    vector<int> t(N+3);
    t[0] = t[1] = 0;
    for(int i=1; i<=N; ++i) {
        int tmp; cin >> tmp;
        t[i+1] = t[i] + tmp;
    }
    t[N+2] = t[N+1];
    vector<int> v(N+2);
    v[0] = 0;
    for(int i=1; i<=N; ++i) cin >> v[i];
    v[N+1] = 0;

    int tmax = t[N+2];
    vector<double> v2(2 * tmax + 1, INF);
    for(int i=0; i<=N+1; ++i) {
        for(int tm=0; tm<=2*tmax; ++tm) {
            if(0.5 * tm < t[i]) {
                v2[tm] = min(v2[tm], v[i] + t[i] - 0.5 * tm);
            } else if(0.5 * tm > t[i+1]) {
                v2[tm] = min(v2[tm], v[i] + 0.5 * tm - t[i+1]);
            } else {
                v2[tm] = min(v2[tm], (double)v[i]);
            }
        }
    }

    double ans = 0;
    for(int tm=0; tm<2*tmax; ++tm) {
        ans += 0.25 * (v2[tm] + v2[tm+1]);
    }
    printf("%.4f\n", ans);
    return 0;
}