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 <stdio.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(), (v).end()

typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, P> PP;

int n,t[102];
double ans,dp[40005],v[102];

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

	REP(i,40005) dp[i]=INF;

	cin>>n;
	REP(i,n) cin>>t[i+1];
	REP(i,n) cin>>v[i+1];
	int tm=0;
	REP(i,n+2) {
		t[i]*=2;
		FOR(j,tm,tm+t[i]+1) {
			dp[j]=min(dp[j],v[i]);
		}
		tm+=t[i];
	}
	FOR(i,1,tm+1) {
		dp[i]=min(dp[i],dp[i-1]+0.5);
	}
	for(int i=tm-1;i>=0;i--) {
		dp[i]=min(dp[i],dp[i+1]+0.5);
	}
	REP(i,tm) {
		ans+=(dp[i]+dp[i+1])*0.25;
	}
	printf("%.10f\n",ans);

	return 0;
}