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>
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
using namespace std;
typedef long long ll;
int INF = (1LL << 30) - 1;
int MOD = 1e9+7;
int main(){
    int N;
    cin >> N;
    vector<int> V(N+2), T(N+2), accumT(N+2);
    rep(i,1,N+1){
        cin >> T[i];
        T[i] *= 2;
        accumT[i+1] = accumT[i] + T[i];
    }
    rep(i,1,N+1){
        cin >> V[i];
        V[i] *= 2;
    }
    vector<int> p(40010);
    rep(i,0,accumT[N+1]+1){
        rep(j,1,N+2){
            if(accumT[j] >= i){
                p[i] = V[j-1];
                break;
            }
        }
        rep(j,1,N+2){
            int d = abs(accumT[j] - i);
            p[i] = min(p[i], d + min(V[j-1], V[j]));
        }
    }
    int sum = 0;
    rep(i,0,accumT[N+1]+1)sum += p[i];
    printf("%.9lf\n", 0.25 * sum);
}