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>
#include <algorithm>
#include <queue>
using namespace std;
int main() {
	int N; 
	cin >> N; 
	int T[110] = {}; 
	int V[110] = {}; 
	int S[110] = {}; 
	for(int i = 1; i <= N; i++){
		cin >> T[i];
		S[i] = S[i-1] + T[i] * 2; 
	}
	for(int i = 1; i <= N; i++){
		cin >> V[i]; 
	}
	int A[40010] = {}; // N+1個の関数の最小値を各点について考える．0.5の整数倍で切り替わることがあるので，あらかじめVもTも2倍しておいて，あとで積分値を4で割る．
	for(int t = 0; t <= S[N]; t++){
		A[t] = min(t, S[N] - t); 
	}
	for(int i = 1; i <= N; i++){
		for(int t = 0; t <= S[N]; t++){
			if(t < S[i-1]){A[t] = min(A[t], V[i] * 2 - t + S[i-1]);}
			else if(t >= S[i-1] && t <= S[i]){A[t] = min(A[t], V[i] * 2);}		
			else if(t > S[i]){A[t] = min(A[t], V[i] * 2 + t - S[i]);}
		}
	}
	int ans = 0; 
	for(int i = 1; i < S[N]; i++){
		ans += A[i]; 
	}
	cout.precision(10); 
	cout << (double) ans * 0.25 << endl; 
}