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;

#define rep(i,j,n) for(int i=(j);i<(n);i++)
#define all(i) i.begin(),i.end()
#define rall(i) i.rbegin(),i.rend()
#define INF 1e9
#define LINF 1e18
const int mod = 1e9+7;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef pair<int, int> pi;
typedef long long i64;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  vi t(n), v(n);
  int sum = 0;
  rep(i, 0, n) {
    cin >> t[i];
    t[i] *= 2;
    sum += t[i];
  }
  rep(i, 0, n) {
    cin >> v[i];
    v[i] *= 2;
  }

  int nw = 0, tot = 0;
  vi d(sum + 1);
  rep(i, 0, n) {
    tot += t[i];
    d[nw] = min(d[nw], v[i]);
    nw++;
    for(; nw <= tot; ++nw) d[nw] = min(d[nw - 1] + 1, v[i]);
    nw--;
  }

  int s = 0, mn = 0;
  for(int i = nw - 1; i > 0; --i) {
    mn = min(mn + 1, d[i]);
    s += mn;
  }
  cout << fixed << setprecision(20) << (double)s / 4 << endl;
}