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 Python (3.8.2) (judge verdict: Accepted)

from math import *
from bisect import bisect_left

def cin(): # To take limited number of inputs
    return map(int,input().split())

def cins(): # To take space sepreated strings
    return input.split()

def cino(test=False): # To take individual int input (test = False)
    if not test:
        return int(input())
    else: # To take string input (test = True)
        return input()

def cina(): # array input
  return list(map(int,input().split()))

def ssplit(): # multiple string input
    return list(input().split())

def printlist(l): # To print space seperated array
    for i in l:
        print(i,end=" ")

def bs(arr, l, r, x): 
    while l <= r: 
        mid = l + (r - l) // 2; 
        if arr[mid] == x: 
            return mid 
        elif arr[mid] < x: 
            l = mid + 1
        else: 
            r = mid - 1
    return -1


def main():
    a,b = cin()
    arr = cina()
    arr.sort()
    ans = 0
    for i in range(b):
        ans += arr[i]
    print(ans)





if __name__ == '__main__':
    main()