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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define rep(i,n) for(i=0;i<n;i++)
#define PI 3.14159265358979323846264338327950L
  /* 配列の要素を交換する */
void Swap(long int x[ ], int i, int j)
{
    int temp;

    temp = x[i];
    x[i] = x[j];
    x[j] = temp;
}
void Qdsort(long int x[ ], int left, int right)
{
    int i, j;
    int pivot;

    i = left;                      /* ソートする配列の一番小さい要素の添字 */
    j = right;                     /* ソートする配列の一番大きい要素の添字 */

    pivot = x[(left + right) / 2]; /* 基準値を配列の中央付近にとる */

    while (1) {                    /* 無限ループ */

        while (x[i] > pivot)       /* pivot より小さい値が */
            i++;                   /* 出るまで i を増加させる */

        while (pivot > x[j])       /* pivot より大きい値が */
            j--;                   /*  出るまで j を減少させる */
        if (i >= j)                /* i >= j なら */
            break;                 /* 無限ループから抜ける */

        Swap(x, i, j);             /* x[i] と x[j]を交換 */
        i++;                       /* 次のデータ */
        j--;
    }
    if (left < i - 1)              /* 基準値の左に 2 以上要素があれば */
        Qdsort(x, left, i - 1);     /* 左の配列を Q ソートする */
    if (j + 1 <  right)            /* 基準値の右に 2 以上要素があれば */
        Qdsort(x, j + 1, right);    /* 右の配列を Q ソートする */
}
int main(){
    int n,k,i;
    scanf("%d%d",&n,&k);
    long int h[n],sum=0;
    rep(i,n){
        scanf("%ld",&h[i]);
        sum+=h[i];
    }
    if(n<=k){
        printf("0");
    }else{
        Qdsort(h,0,n-1);
        rep(i,k) sum=sum-h[i];
        printf("%ld",sum);
    }
    return 0;
}