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<cstdio>
    #include<algorithm>
    using namespace std;
     
    int n, len;
    int t[111], v[111];
    double ma[111*222*2];
     
    int main() {
    	scanf("%d", &n);
    	len = 0;
    	for(int i=1;i<=n;i++) {
    		scanf("%d", &t[i]);
    		len += t[i]*2;
    	}
    	int it = 0;
    	ma[0] = 0;
    	for(int i=1;i<=n;i++) {
    		scanf("%d", &v[i]);
    		for(int j=0;j<t[i]*2;j++) {
    			it++;
    			ma[it] = min(1.*ma[it-1]+.5, 1.*v[i]);
    		}
    	}
    	it = len;
    	ma[len] = 0;
    	for(int i=n;i>0;i--) {
    		for(int j=0;j<t[i]*2;j++) {
    			it--;
    			ma[it] = min(ma[it], min(1.*ma[it+1]+.5, 1.*v[i]));
    		}
    	}
    	
    	double ans = 0LL;
    	for(int i=1;i<=len;i++)
    		ans += ma[i-1] + ma[i];
    	printf("%.12lf\n", ans/4);
    	
    	return 0;
    }