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,n) for(long long i = 0; i < (long long)(n); i++)
#define repi(i,a,b) for(long long i = (long long)(a); i < (long long)(b); i++)
template <typename T> istream& operator>>(istream& i, vector<T>& o) { rep(j, o.size()) i >> o[j]; return i;}

const int tres = 4, vres = 4;
vector<int> dp(20010*tres), lim(20010*tres);
int main(void) {
    int n; cin >> n;
    vector<int> t(n), v(n); cin >> t >> v;
    rep(i, n) v[i] *= vres, t[i] *= tres;
    int s = 0;
    rep(i, n) {
        repi(j, s, s+t[i]) 
            lim[j] = v[i];
        s += t[i];
    }
    
    rep(_, 210*vres) repi(i, 1,dp.size()) 
        if (dp[i] <= min({dp[i+1], dp[i-1], lim[i] -1, lim[i-1]-1}))
            dp[i]++;

    double ret = 0;
    rep(i, dp.size()-1) 
        ret += (dp[i] + dp[i+1]) / (double)vres * (1.0 / (double(tres))) / 2.0;

    cout << setprecision(40) << ret << endl;
    return 0;
}