You are given the array a consisting of n elements and the integer k ≤ n.

You want to obtain at least k equal elements in the array a. In one move, you can make one of the following two operations:

  * Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of a is mn then you choose such index i that a_i = mn and set a_i := a_i + 1); 
  * take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of a is mx then you choose such index i that a_i = mx and set a_i := a_i - 1). 



Your task is to calculate the minimum number of moves required to obtain at least k equal elements in the array.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a and the required number of equal elements.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of moves required to obtain at least k equal elements in the array.

Examples

Input


6 5
1 2 2 4 2 3


Output


3


Input


7 5
3 3 2 1 1 1 3


Output


4
INF = int (2e9)
n, k = list (map (int, input ().split ()))
a = list (map (int, input ().split ()))
a.sort()
a = [(x, 0) for x in a]

ans = INF
ha = {}
tail = n - 1
p = 0

def lowerbound (x):
    l, r = 0, tail - 1
    while (l <= r) :
        mid = (l + r) >> 1
        if (a[mid][0] >= x) :
            r = mid - 1
        else :
            l = mid + 1
    return r + 1

def find (x):
    l = lowerbound (x[0])
    r = lowerbound (x[0] + 1) - 1
    while (l <= r) :
        mid = (l + r) >> 1
        if (a[mid][1] <= x[1]) :
            r = mid - 1
        else :
            l = mid + 1
    return r + 1

while (a[tail][0]):
    flg = True
    cnt = 0
    val = a[tail][0]

    if (a[tail - k + 1][0] == val) :
        for j in range(k) :
            cnt += a[tail - j][1]
        ans = min (ans, cnt)

    rem = k
    while (a[tail][0] == val) :
        if (rem > 0) :
            tup = (a[tail][0] // 2, a[tail][1] + 1)
            a.pop ()
            a.insert(find (tup), tup)
            rem -= 1
            # print (a)
        else :
            a.pop ()
            tail -= 1

print (ans)
	    		  				 	  				 	 			  	