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>
#include <cstdio>
using namespace std;

const int MAX_N = (int)110;

int n, t[MAX_N], v[MAX_N];
int up[400 * 100 + 10] = {};
int dp[400 * 100 + 10] = {};

int main(void) {
    // Here your code !
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &t[i]);
    }

    int ind = 0;
    for (int i = 0; i < n; ++i) {
      scanf("%d", &v[i]);
      for (int j = 0; j < 2 * t[i]; ++j) {
        up[ind] = v[i] * 2;
        ind++;
      }
    }

    int tmp;
    for (int i = 0; i < 210; ++i) {
      for (int j = 1; j < ind + 1; ++j) {
        tmp = min({dp[j - 1], dp[j + 1], up[j] - 1, up[j - 1] - 1});
        if (dp[j] <= tmp) dp[j]++;
      }
    }

    double whole = 0;
    for (int i = 0; i < ind + 1; ++i) {
      whole += (dp[i] + dp[i + 1]) * 0.5;
    }

    printf("%f\n", whole * 0.25);

    return 0;
}