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 JAVA):

Solution:
import java.util.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int[] t = new int[n];
		int[] v = new int[n];
		int sumT = 0;
		for(int i = 0; i < n; i++){
		    t[i] = sc.nextInt();
		    sumT += t[i];
		}
		for(int i = 0; i < n; i++){
		    v[i] = sc.nextInt();
		}
		
		// 計算
		double result = 0;
		int[] V = new int[sumT];
		int[] minV = new int[sumT+1];
		int index = 0;
		for(int i = 0; i < n; i++){
		    for(int j = 0; j < t[i]; j++){
		        V[index] = v[i];
		        index++;
		        if(index == sumT) continue;
		        if(j < t[i]-1){
		            minV[index] = v[i];
		        }else{
		            minV[index] = min(v[i], v[i+1]);
		        }
		    }
		}
		int[] minA = new int[sumT+1];
		for(int i = 0; i < (sumT+2)/2; i++){
		    minA[i] = i;
		    minA[sumT-i] = i;
		}
		int[] r = new int[sumT+1];
		for(int i = 1; i < sumT+1; i++){
		    r[i] = min(r[i-1]+1, min(minV[i], minA[i]));
		}
		int[] l = new int[sumT+1];
		for(int i = 1; i < sumT+1; i++){
		    l[sumT-i] = min(l[sumT-i+1]+1, min(minV[sumT-i], minA[sumT-i]));
		}
		int[] res = new int[sumT+1];
		for(int i = 0; i < sumT+1; i++){
		    res[i] = min(r[i], l[i]);
		}
		
		for(int i = 1; i < sumT+1; i++){
		    result += (double)abs(res[i]-res[i-1])/2 + min(res[i], res[i-1]);
		}
		for(int i = 1; i < sumT+1; i++){
		    if(res[i] == res[i-1] && res[i-1] < V[i-1]) result += 0.25;
		}
		//System.out.println(Arrays.toString(V));
		//System.out.println(Arrays.toString(minV));
		//System.out.println(Arrays.toString(minA));
		//System.out.println(Arrays.toString(r));
		//System.out.println(Arrays.toString(l));
		//System.out.println(Arrays.toString(res));
		
		// 出力
		System.out.println(result);
	}
}