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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

#define sz(x) ((int)(x).size())
#define in(x) int x;cin>>x
#define inn(x, y) int x,y;cin>>x>>y
#define innn(x, y, z) int x,y,z;cin>>x>>y>>z
#define bit(n) (1<<(n))

#define out(x) cout<<(x)<<endl;
#define err(x) cerr<<(x)<<endl;
#define errr(x,y) cerr<<(x)<<" "<<(y)<<endl;
#define errrr(x,y,z) cerr<<(x)<<" "<<(y)<<" "<<(z)<<endl;

#define rep(i,n) for (int i=0;i<(n);i++)
#define repp(i,n) for (int i=1;i<=(n);i++)
#define reppp(j, i, n) for (int j=i+1; j<n; j++)

#define lim5 200002
#define lim9 1000000002

#define deg(rad) (((rad)/2/M_PI)*360)
#define mid(min, max) ((min + max) / 2)

int main() {
	inn(n, k);
	ll h[lim5] = {};

	rep(i, n) {
		cin >> h[i];
	}

	sort(h, h + n);

	ll cnt = 0;
	rep (i, n - k) {
		cnt += h[i];
	}

	out(cnt);

	return 0;
}