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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ld = long double;
using P = pair<int,int>;

int main() {
  int n;
  cin >> n;
  vector<ld> t(n),v(n);
  rep(i,n) cin >> t[i];
  rep(i,n) cin >> v[i];

  vector<ld> m(n);
  rep(i,n-1){
    int j = n-1-i;
    m[j-1] = min(m[j]+t[j],v[j]);
  }

  ld ans = 0;
  ld vl = 0;
  rep(i,n){
    ld vr = min({vl+t[i],v[i],m[i]});
    ld tl = max(ld(0),v[i]-vl);
    ld tr = max(ld(0),v[i]-vr);
    if(tl+tr>t[i]){
      ld d = (t[i]-tl-tr)/2;
      tl -= d;
      tr -= d;
      if(tl<0) tr += tl, tl = 0;
      if(tr<0) tl += tr, tr = 0;
    }
    ans += (vl+vl+tl)*tl/2 
        + (vr+vr+tr)*tr/2 
        + (t[i]-tl-tr)*v[i];
    vl = vr;
  }
  printf("%.10Lf\n",ans);
}