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: Accepted)


#include<stdio.h>


#define SWAP(type,a,b) { type work = a; a = b; b = work; }

/*
    基準値を選ぶ
*/
long int select_pivot(long  int* array, int begin, int end)
{
    return array[(begin + end) / 2];  // 中央の要素を基準値とする
}


/*
    クイックソート (再帰部分、昇順)
*/
void quick_sort_rec(long int* array, int begin, int end)
{
    long int pivot = select_pivot( array, begin, end );
    int i = begin;
    int j = end;

    while( 1 ){
        while( array[i] < pivot ){ ++i; }  // 基準値以上の値が見つかるまで右方向へ進めていく
        while( array[j] > pivot ){ --j; }  // 基準値以下の値が見つかるまで左方向へ進めていく

        if( i >= j ){ break; }  // 左右から進めてきたiとjがぶつかったらループを終える

        // 基準値の位置よりも左側にあり、基準値よりも大きい値 (array[i]) と、
        // 基準値の位置よりも右側にあり、基準値よりも小さい値 (array[j]) の
        // 位置関係を交換する。
        SWAP( int, array[i], array[j] );

        // 次回に備えて、注目点をずらす
        i++;
        j--;
    }


    // 基準値の位置より左側に要素が２つ以上あれば、
    // その部分についてクイックソートを再帰させる
    if( i - begin >= 2 ){
        quick_sort_rec( array, begin, i - 1 );
    }

    // 基準値の位置より右側に要素が２つ以上あれば、
    // その部分についてクイックソートを再帰させる
    if( end - j >= 2 ){
        quick_sort_rec( array, j + 1, end );
    }
}


void sort(long int* array, int size){
    // 配列全体を対象にする
    quick_sort_rec( array, 0, (int)(size - 1) );
}

void print_array(long int data[], int n){
    for(int i = 0; i < n; i++){
        printf("%ld ", data[i]);
    }
    printf("\n");
}

int main(void){
    int n, k;
    scanf("%d""%d", &n, &k);
    long int h[n], cnt = 0;;
    for(int i = 0; i < n; i++){
        scanf("%ld", &h[i]);
    }
    sort(h, n);
    for(int i = n - k - 1; i >= 0; i--){
        cnt += h[i];
    }
    printf("%ld\n", cnt);
    return 0;
}