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>
#include <stdlib.h>
void quick_sort( int h[], int right, int left );

void swap( int *a, int *b ){
    int temp = 0;
    temp = *a;
    *a = *b;
    *b = temp;
    }

int main(void){
    // 変数宣言
    int n, k;
    int *h;
    h = (int *)malloc( sizeof(int) * n );
    int i;
    int result = 0;
    unsigned long total = 0;

    // 入力処理
    scanf( "%d\n", &n );
    scanf( "%d\n", &k );
    for( i = 0 ; i < n ; i++ ){
        scanf( "%d", &h[ i ] );
    }

    // 演算処理
    if( n > k ){
        // モンスターの体力をソートする
        quick_sort( h, ( n - 1 ), 0 );

        // モンスター数 - kの回数
        n = n - k ;

        // モンスターHPの総和
        for( i = 0; i < n ; i++){
            total += h[i];
        }

    }else{
        total = 0;
    }
    // 出力
    printf("%lu", total );

    return 0;
}

void quick_sort( int h[], int right, int left ){
    // 変数宣言
    int i;
    int pivot;
    int Right = right, Left = left;

    // 中央値を決める
    pivot = h[( right + left ) / 2 ];

    while( 1 ){
        // 中央値を右から比較
        while( pivot < h[ Right ] ){
            Right--;
        }

        // 中央値を左から比較
        while( h[ Left ] < pivot ){
            Left++;
        }

        // カーソルが重なったらループを抜ける
        if( Right <= Left ){
            break;
        }

        // 入れ替え処理
        swap( &h[Right], &h[Left] );
        Right--;
        Left++;
    }
    // ソートを続けるか
    if( left < ( Left - 1 ) ){
        quick_sort( h, Left - 1, left );
    }

    if( ( Right + 1 < right ) ){
        quick_sort( h, right, Right + 1 );
    }

}