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 <cstdio>
#include <algorithm>
using namespace std;
int t[105];
double v[105];
double u[40005];
int main( void )
{
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  t[0] = 0;
  v[0] = 0;
  v[N+1] = 0;
  for( int i=1; i<=N; i++ ) cin >> t[i];
  for( int i=1; i<=N; i++ ) cin >> v[i];
  for( int i=1; i<=N; i++ ){
    t[i] += t[i-1];
  }
  for( int i=0; i<=t[N]*2; i++ ){
    u[i] = 0.5 * i;
  }
  for( int i=t[N]*2; i>=0; i-- ){
    u[i] = min( u[i], t[N] - i*0.5 );
  }
  for( int i=1; i<=N; i++ ){
    for( int j=0; j<t[i-1]*2; j++ ){
      u[j] = min( u[j], v[i] + t[i-1] - j*0.5 );
    }
    for( int j=t[i-1]*2; j<=t[i]*2; j++ ){
      u[j] = min( u[j], v[i] );
    }
    for( int j=t[i]*2+1; j<=t[N]*2; j++ ){
      u[j] = min( u[j], v[i] + j*0.5 - t[i] );
    }
  }
  double ret = 0;
  for( int i=1; i<=t[N]*2; i++ ){
    ret += (u[i-1] + u[i]) / 4;
  }
  printf( "%.12f\n", ret );
}