You are given an array a_1, a_2, ..., a_n and an integer k.

You are asked to divide this array into k non-empty consecutive subarrays. Every element in the array should be included in exactly one subarray. Let f(i) be the index of subarray the i-th element belongs to. Subarrays are numbered from left to right and from 1 to k.

Let the cost of division be equal to ∑_{i=1}^{n} (a_i ⋅ f(i)). For example, if a = [1, -2, -3, 4, -5, 6, -7] and we divide it into 3 subbarays in the following way: [1, -2, -3], [4, -5], [6, -7], then the cost of division is equal to 1 ⋅ 1 - 2 ⋅ 1 - 3 ⋅ 1 + 4 ⋅ 2 - 5 ⋅ 2 + 6 ⋅ 3 - 7 ⋅ 3 = -9.

Calculate the maximum cost you can obtain by dividing the array a into k non-empty consecutive subarrays. 

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3 ⋅ 10^5).

The second line contains n integers a_1, a_2, ..., a_n ( |a_i| ≤ 10^6). 

Output

Print the maximum cost you can obtain by dividing the array a into k nonempty consecutive subarrays. 

Examples

Input


5 2
-1 -2 5 -4 8


Output


15


Input


7 6
-3 0 -1 -2 -2 -4 -1


Output


-45


Input


4 1
3 -1 6 0


Output


8
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7

from fractions import Fraction
import heapq as hp
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()



n,k = Ri()
a= Ri()

suffix = [0]*n
suffix[n-1] = a[n-1]
for i in range(len(a)-2,-1,-1):
    suffix[i] = suffix[i+1]+a[i]
suffix = suffix[1:]
suffix.sort(reverse = True)
summ  = sum(a)
for i in range(k-1):
    summ+=suffix[i]
print(summ)



