Problem:
Score : 300 points

Problem Statement

Fennec is fighting with N monsters.

The health of the i-th monster is H_i.

Fennec can do the following two actions:

Attack: Fennec chooses one monster. That monster's health will decrease by 1.

Special Move: Fennec chooses one monster. That monster's health will become 0.

There is no way other than Attack and Special Move to decrease the monsters' health.

Fennec wins when all the monsters' healths become 0 or below.

Find the minimum number of times Fennec needs to do Attack (not counting Special Move) before winning when she can use Special Move at most K times.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 2 \times 10^5

1 \leq H_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
H_1 ... H_N

Output

Print the minimum number of times Fennec needs to do Attack (not counting Special Move) before winning.

Sample Input 1

3 1
4 1 5

Sample Output 1

5

By using Special Move on the third monster, and doing Attack four times on the first monster and once on the second monster, Fennec can win with five Attacks.

Sample Input 2

8 9
7 9 3 2 3 8 4 6

Sample Output 2

0

She can use Special Move on all the monsters.

Sample Input 3

3 0
1000000000 1000000000 1000000000

Sample Output 3

3000000000

Watch out for overflow.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
//ソート関数がTLEするよ
#include <stdio.h>
#include <string.h>
#define len(arr) sizeof(arr) / sizeof(*arr)
void swap(int *a, int *b) {
    int tmp = *a;
    *a = *b;
    *b = tmp;
}

int max(int a, int b){
    if (a >= b){
        return a;
    }
    else {
        return b;
    }
}

int partition(int *arr, int left, int right){
    int pivod_num = left;
    int pivod_element = arr[pivod_num];

    int left_num = left;
    int right_num = right;
    int next_first;
    while (left_num < right_num){
        int swap_flag = 1;
        if (arr[left_num] < pivod_element){
            left_num += 1;
            swap_flag = 0;
        }
        if (arr[right_num] >= pivod_element){
            right_num -= 1;
            swap_flag = 0;
        }
        if (swap_flag){
            swap(&arr[left_num], &
            arr[right_num]);
        }
    }

    if (left_num == right_num){
        if (arr[left_num] >= pivod_element){
            next_first = max(left, left_num - 1);
        }
        else {
            next_first = left_num;
        }
    }
    else {
        next_first = right_num;
    }

    return next_first;
}

void sort(int *arr, int left, int right){
    int left_first = left;
    int right_latter = right;
    int right_first = partition(arr, left, right);
    int left_latter = right_first + 1;
    //printf("%d %d\n", left, right);
    //for (int i = 0; i < 10; i++){
    //    printf("%d ", arr[i]);
    //}
    if (left_first < right_first){
        sort(arr, left_first, right_first);
    }
    if (left_latter < right_latter){
        sort(arr, left_latter, right_latter);
    }
}

void quick_sort(int *arr, int size){
    int left = 0;
    int right = size - 1;
    //printf("%d %d\n", left, right);
    sort(arr, left, right);
}



int main(void){
    int n, k;
    scanf("%d%d", &n, &k);
    int h[n];
    for (int i = 0; i < n; i++){
        scanf("%d", &h[i]);
    }

    int ans;
    if (k >= n){
        ans = 0;
    }
    else {
        quick_sort(h, n);
        int sum = 0;
        for (int i = 0; i < n - k; i++){
            sum += h[i];
        }
        ans = sum;
    }
    printf("%d\n", ans);

    return 0;
}