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;

long long int N, t[114], v[114], R[252525], a, i;

int main(){
    scanf("%lld", &N);
    *t = 0;
    *v = 0;
    for(; i < N; ++i){scanf("%lld", t + i + 1);t[i + 1] *= 2;}
    for(i = 0; i < N; ++i){scanf("%lld", v + i + 1);v[i + 1] *= 2;}
    t[N + 1] = 0;
    v[N + 1] = 0;
    partial_sum(t, t + N + 2, t);
    for(i = 1; i <= t[N + 1]; ++i){
        R[i] = 1145141919810LL;
        for(int j = 0; j <= N + 1; ++j){
            R[i] = min(abs(min({t[j] - i, 0LL, j ? i - t[j - 1] : 0LL})) + v[j], R[i]);
        }
    }
    for(i = 1; i <= t[N + 1]; ++i){
        a += (R[i] + R[i - 1]);
    }
    a >>= 1;
    printf("%lld", a / 4);
    switch(a & 3){
    case 1:
        puts(".25");
        break;
    case 2:
        puts(".5");
        break;
    case 3:
        puts(".75");
        break;
    }
    return 0;
}