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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=1LL<<61;

int main() {
  ll n;
  cin >> n;
  vec t(n),v(n);
  for(ll i=0;i<n;i++) cin >> t[i],t[i]*=2;
  for(ll i=0;i<n;i++) cin >> v[i],v[i]*=2;
  for(ll i=0;i<n-1;i++) t[i+1]+=t[i];
  ll ct=0;
  ll cv=0;
  vec lim(8e4+10);
  for(ll i=0;i<n;i++) {
    if(i) ct--,cv--;
    while(ct<=t[i]) {
      lim[ct]=min(cv,v[i]);
      cv=lim[ct]+1;
      ct++;
    }
  }
  cv=0;
  ct--;
  double ans=0;
  while(ct) {
    cv=min(lim[ct],cv);
    ans+=cv;
    ct--;
    cv++;
  }
  ans*=1/4.0;
  cout << fixed << setprecision(10) << ans << endl;
}