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;
typedef long long ll;
static const int MAX=100;
int g[MAX*400+2];

int n,t[MAX+2],v[MAX+2],ans;
int main(void){
	cin >> n;
	t[0]=0;
	for(int i=1;i<=n;i++){
		cin >>t[i];
		t[i]*= 2;//x->2*x座標にする
		t[i]+=t[i-1];//2*x座標と対応させる
	}
	for(int i=0;i<n;i++){
		cin >>v[i];
		v[i]*= 2;//y座標も2倍。傾きが1のまま。
	}
	for(int i=0;i<=t[n];i++){
		g[i]=min(i,t[n]-i);//t[0],t[n]の制限
		for(int j=0;j<n;++j){
			if(i<=t[j]){//i->t[j]->t[j+1]
				g[i]=min(g[i],v[j]+t[j]-i);
			}else if(t[j+1]<=i){//t[j]->t[j+1]->i
				g[i]=min(g[i],v[j]+i-t[j+1]);
			}else{//t[j]->j->t[j+1]
				g[i]=min(g[i],v[j]);
			}
		}
	}
	for(int i=0; i<t[n]; ++i){
		ans += g[i]+g[i+1];//台形の面積
	}
	printf("%.5lf\n",((double)(ans)/8));
	return 0;
}