And again a misfortune fell on Poor Student. He is being late for an exam.

Having rushed to a bus stop that is in point (0, 0), he got on a minibus and they drove along a straight line, parallel to axis OX, in the direction of increasing x.

Poor Student knows the following: 

  * during one run the minibus makes n stops, the i-th stop is in point (xi, 0)
  * coordinates of all the stops are different 
  * the minibus drives at a constant speed, equal to vb
  * it can be assumed the passengers get on and off the minibus at a bus stop momentarily 
  * Student can get off the minibus only at a bus stop 
  * Student will have to get off the minibus at a terminal stop, if he does not get off earlier 
  * the University, where the exam will be held, is in point (xu, yu)
  * Student can run from a bus stop to the University at a constant speed vs as long as needed 
  * a distance between two points can be calculated according to the following formula: <image>
  * Student is already on the minibus, so, he cannot get off at the first bus stop 



Poor Student wants to get to the University as soon as possible. Help him to choose the bus stop, where he should get off. If such bus stops are multiple, choose the bus stop closest to the University.

Input

The first line contains three integer numbers: 2 ≤ n ≤ 100, 1 ≤ vb, vs ≤ 1000. The second line contains n non-negative integers in ascending order: coordinates xi of the bus stop with index i. It is guaranteed that x1 equals to zero, and xn ≤ 105. The third line contains the coordinates of the University, integers xu and yu, not exceeding 105 in absolute value. 

Output

In the only line output the answer to the problem — index of the optimum bus stop.

Examples

Input

4 5 2
0 2 4 6
4 1


Output

3

Input

2 1 1
0 100000
100000 100000


Output

2

Note

As you know, students are a special sort of people, and minibuses usually do not hurry. That's why you should not be surprised, if Student's speed is higher than the speed of the minibus.
import math
def main():
       numberOfBusSTOPS, busSpeed, studentSpeed = map(int,input().split())
       busCoordinate = []
       #the time students needs to go to the university based on the station he left the bus
       studentTime = []
       #the distance from university based on bus stations
       distanceFromUniversity = []
       #the time in which the bus arrives to the station
       busTime = []
       busCoordinate = list(map(int,input().split()))
       #total time bus + student
       time = []
       universityX, universityY = map(int,input().split())
       #we have to compute thetime for each station
       for i in range(numberOfBusSTOPS):
              #according to teh formula t=d/v
              busTime.append(busCoordinate[i]/busSpeed)
              #according to the formula sqrt((x2-x1)^2+(y2-y1)^2)
              distanceFromUniversity.append(((universityX - busCoordinate[i])**2 + universityY**2)**(1/2))
              #according to teh formula t=d/v
              studentTime.append(distanceFromUniversity[i]/studentSpeed)
              #total time is the sum of student time and bus time
              time.append(busTime[i]+studentTime[i])
       #next we have to find the minimum value for time and its index, because we will have to return the index
       minTime = min(time)
       result = time.index(min(time))
       #we have to check if there exists another time == minTime, but the station is closer to the university
       for i in range(numberOfBusSTOPS):
              #we check if they are equal substracting them, because they might differ some digits because of computer arithmetic
              if abs(minTime - time[i]) < 10**-7 and distanceFromUniversity[i]<distanceFromUniversity[result]:
                     result = i
      #also we have to take into consideration that student cant get off at the first station
      #if the time needed if he get off at first station is the smallest one
      #than he has to get off at the next station
       if result == 0:
              result += 1
       print(result+1)
main()
       
