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
N, K = [int(x) for x in input().split()]
l = [int(x) for x in input().split()]
lk = [0] * (1000000)
for n in l:
	t = n
	while t != 0:
		lk[t] += 1
		t //= 2
lk[0] = 99999999
g = 0
mn = 9999999
for i in range(len(lk) - 1, -1, -1):
	if(lk[i] >= K):
		g = i;
		ed = []
		for n in l:
			if(n < g): ed.append(9999999)
			elif(n == g): ed.append(0)
			else:
				t = n
				c = 0
				while True:
					t //= 2
					c += 1
					if(t <= g):
						if(t == g): ed.append(c)
						else: ed.append(9999999)
						break
		ed.sort()
		mn = min(mn, sum(ed[:K]))
print(mn)
# ed = []
# for n in l:
# 	if(n < g): ed.append(9999999)
# 	elif(n == g): ed.append(0)
# 	else:
# 		t = n
# 		c = 0
# 		while True:
# 			t //= 2
# 			c += 1
# 			if(t <= g):
# 				if(t == g): ed.append(c)
# 				else: ed.append(9999999)
# 				break
# ed.sort()
# print(sum(ed[:K]))
