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>
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const double INF=1e9;

int main(){
  int n;
  cin>>n;
  vector<int>t(n),v(n);
  int T=0;
  rep(i,n){
    cin>>t[i];
    T+=t[i];
  }
  rep(i,n)cin>>v[i];

  //初期化
  vector<double>maxV(2*T+1,INF);
  int now=0;
  rep(i,n){
    rep(j,2*t[i]){
      chmin(maxV[now],(double)v[i]);
      chmin(maxV[now+1],(double)v[i]);
      now++;
    }
  }

  //各点の位置を決める
  maxV[0]=maxV[2*T]=0;

  rep(i,2*T){
    chmin(maxV[i+1],maxV[i]+0.5);
  }

  for(int i=2*T; i>0; --i){
    chmin(maxV[i-1],maxV[i]+0.5);
  }

  //台形の面積を計算
  double ans=0;
  rep(i,2*T){
    ans+=(maxV[i]+maxV[i+1])*0.5/2;
  }
  printf("%.6f\n",ans);
}