Problem:
Score : 200 points

Problem Statement

A shop sells N kinds of fruits, Fruit 1, \ldots, N, at prices of p_1, \ldots, p_N yen per item, respectively. (Yen is the currency of Japan.)

Here, we will choose K kinds of fruits and buy one of each chosen kind. Find the minimum possible total price of those fruits.

Constraints

1 \leq K \leq N \leq 1000

1 \leq p_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
p_1 p_2 \ldots p_N

Output

Print an integer representing the minimum possible total price of fruits.

Sample Input 1

5 3
50 100 80 120 80

Sample Output 1

210

This shop sells Fruit 1, 2, 3, 4, and 5 for 50 yen, 100 yen, 80 yen, 120 yen, and 80 yen, respectively.

The minimum total price for three kinds of fruits is 50 + 80 + 80 = 210 yen when choosing Fruit 1, 3, and 5.

Sample Input 2

1 1
1000

Sample Output 2

1000

Solution in PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

import itertools
import sys
from typing import Callable, NoReturn, Tuple


def main() -> NoReturn:
    readline: Callable[[], str] = sys.stdin.readline
    n, k = map(int, readline().rstrip().split())  # type: int,...
    p: Tuple[int] = tuple(int(_) for _ in readline().rstrip().split())

    if k < n // 2:
        answer: int = 10 ** 6
        for combinations in itertools.combinations(p, k):
            sum_of_comb: int = sum(combinations)
            if sum_of_comb < answer:
                answer = sum_of_comb
    else:
        not_chosen: int = 0
        for combinations in itertools.combinations(p, n - k):
            sum_of_comb: int = sum(combinations)
            if sum_of_comb > not_chosen:
                not_chosen = sum_of_comb
        answer: int = sum(p) - not_chosen

    print(answer)


if __name__ == '__main__':
    main()