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;
const int MOD=1e9+7;
const int INF=1e9;

int dp[40010][202];

int main(){
    int n;
    cin>>n;
    int t[n],v[n];
    for(int i=0;i<n;i++){
        cin>>t[i];
    }
    for(int i=0;i<n;i++){
        cin>>v[i];
    }
    int lmt[40010]={},crt=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<2*t[i];j++){
            lmt[++crt]=2*v[i];
        }
    }
    for(int i=0;i<40010;i++){
        for(int j=0;j<202;j++){
            dp[i][j]=-INF;
        }
    }
    dp[0][0]=0;
    for(int i=1;i<40010;i++){
        for(int j=0;j<=lmt[i];j++){
            if(j>0)dp[i][j]=max(dp[i][j],dp[i-1][j-1]+2*j+1);
            if(j<lmt[i])dp[i][j]=max(dp[i][j],dp[i-1][j+1]+2*j-1);
            dp[i][j]=max(dp[i][j],dp[i-1][j]+2*j);
        }
    }
    cout<<fixed<<setprecision(5)<<dp[40009][0]/8.0<<endl;
}