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: Wrong Answer)

//保留所有权利。
#include<iostream>
#include<stdio.h>//printするよ
#include<vector>
#include<algorithm>//辞書順はnext_permutationだよ
#include<set>
#include<string>
#include<string.h>
#include<time.h>
#include<random>
#include<cmath>
#include<math.h>
#include<queue>
#include<time.h>
#include<iomanip>
#include<utility>//swapで数値交換するよ
//END OF LIBRARIES.
#define rt "\n"//改行を最速化させるよ
#define rep(i,n) for(int i=0;i<n;i++)
#define rop(i,n) for(int i=1;i<=n;i++)
#define drep(i,n) for(int i=n-1;0<=i;i--)
#define yes(ans) if(ans)cout<<"yes"<<rt;else cout<<"no"<<rt;
#define Yes(ans) if(ans)cout<<"Yes"<<rt;else cout<<"No"<<rt;
#define YES(ans) if(ans)cout<<"YES"<<rt;else cout<<"NO"<<rt;
#define vsort(data) sort(data.begin(),data.end(),greater<int>());
#define vcin(v) for (int i = 0; i < (v).size(); i++) { cin >> (v)[i]; }
#define reve(s) reverse(s.begin(),s.end())
#define asas int ans=0
#define cncn int cnt=0
#define smsm int sum=0
#define str srting
#define nn int n=0
#define please return
#define AC 0//おまじないだよ
#define Rapid cin.tie(0);ios::sync_with_stdio(false)
//END OF DEFINE.
using namespace std;
typedef vector<int> vint;
typedef vector<string> vstr;
typedef vector<char> vcha;
typedef vector<double> vdou;
typedef long long int llint;
typedef pair<int, int> pint;
typedef pair<llint, llint> pllint;
typedef vector<llint> vllint;
typedef vector<pint> vpint;
typedef vector<pair<llint, llint>> vpllint;
typedef vector<vector<int>> vvint;
typedef vector<vector<llint>> vvllint;
typedef vector<vector<pint>> vvpint;
typedef vector<bool> vbool;
//END OF TYPEDEF.

//最大公約数（GCD）を求めるよ
//最小公倍数（LCM）は＜＜　A × B ÷ GCD　 ＞＞で求まるよ

long long GCD(long long a, long long b) {
	if (b == 0) return a;
	else return GCD(b, a % b);
}
long long LCM(long long a, long long b) {
	return a * b / GCD(a, b);
}
int x = 0, y = 0;
//BEGIN OF CODE.
int main(void) {
	int n, k;
	cin >> n >> k;
	vint h(n);
	rep(i, n)cin >> h[i];
	sort(h.begin(), h.end(), greater<int>());
	cncn;
	rep(i, n) {
		if (k != 0) {
			h[i] = 0;
			k--;
		}
		else {
			if (h[i] != 0) {
				cnt+=h[i];
			}
		}
	}
	cout << cnt << rt;
	please AC;
}
//THE END.