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>

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define rep(i,n) FOR(i,0,n)
#define RFOR(i,a,b) for(int i=(a)-1;i>=(b);i--)
#define rrep(i,n) RFOR(i,n,0)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n;
	cin >> n;

	int t[n+3];
	t[0] = 0;
	t[1] = 0;
	rep(i, n) {
		cin >> t[i+2];
		t[i+2] += t[i+1];
	}
	t[n+2] = t[n+1];

	int mt = t[n+1];

	int v[n+2];
	v[0] = 0;
	rep(i, n) cin >> v[i+1];
	v[n+1] = 0;

	double mnv[2*mt+1];
	rep(i, 2*mt+1) mnv[i] = 1000.0;

	rep(i, n+2){
		int l = 2*t[i];
		int r = 2*t[i+1];
		
		rep(j, 2*mt+1){
			if(j < l) mnv[j] = min(mnv[j], 0.5 * (l - j)+ v[i]);
			else if(l <= j && j < r) mnv[j] = min(mnv[j], 1.0 * v[i]);
			else mnv[j] = min(mnv[j], 0.5 * (j - r)+ v[i]);
		}
	}

	double ans = 0.0;
	rep(i,2*mt){
		ans += 0.5 * (mnv[i] + mnv[i+1]) / 2;
	}

	cout << fixed << setprecision(5) << ans << endl;
}