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, n) for(ll i = ((ll) a); i < ((ll) n); i++)
using namespace std;
typedef long long ll;
 
ll N, T[101], V[101];
double dp[20001][101];
 
int main(void) {
  cin >> N;
  REP(i, 0, N) cin >> T[i];
  REP(i, 0, N) cin >> V[i];
 
  T[N] = 0;
  REP(i, 0, N) T[i + 1] += T[i];
  V[N] = 0;
 
  REP(i, 0, 20001) REP(j, 0, 101) dp[i][j] = -1e20;
  dp[0][0] = 0;
  ll vi = 0;
  REP(t, 1, T[N] + 1) {
    ll vmax = V[vi], lvmax = V[vi];
    if(t >= T[vi]) {
      vmax = min(V[vi], V[vi + 1]);
      lvmax = V[vi];
      vi++;
    }
    REP(v, 0, vmax + 1) {
      dp[t][v] = max(dp[t][v], dp[t - 1][v] + v);
      if(v <= vmax && v + 1 <= lvmax) dp[t][v] = max(dp[t][v], dp[t - 1][v] + v + 0.25);
      if(v + 1 <= 100) dp[t][v] = max(dp[t][v], dp[t - 1][v + 1] + v + 0.5);
      if(v - 1 >= 0) dp[t][v] = max(dp[t][v], dp[t - 1][v - 1] + v - 0.5);
    }
  }
  printf("%.15lf\n", dp[T[N]][0]);
}