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<bits/stdc++.h>
using namespace std;

#define fastio 			ios::sync_with_stdio(0);	cin.tie(0); cout.tie(0);  cout<<fixed;  cout<<setprecision(12);
#define randomINT 		mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
#define	newl 			cout<<"\n"
#define DISP(as)		for(auto it : as) cout<<it<<" ";newl;
#define all(x)         	(x).begin(),(x).end()
#define mset(x,val)    	memset(x,val,sizeof(x))
#define newl           	cout<<"\n"
#define pb             	push_back
#define mp             	make_pair
#define f 				first
#define s 				second
#define dline          cerr<<"///REACHED///\n";
#define deb1(x)        cerr<<#x<<" = "<<x<<'\n';
#define deb2(x,y)      cerr<<'['<<#x<<','<<#y<<"] = "<<'['<<x<<','<<y<<']'<<'\n';
#define deb3(x,y,z)    cerr<<'['<<#x<<','<<#y<<','<<#z<<"] = "<<'['<<x<<','<<y<<','<<z<<']'<<'\n';

typedef long long 				ll;
typedef long double 			ld;
typedef vector<ll> 				vll;
typedef pair<ll , ll> 			pll;
typedef pair<ld, ld> 			pld;
typedef unordered_map<ll, ll> 	um;
typedef vector<pll> 			vpll;

const ll 	MAX5 	= 	1e+5 + 7;
const ll 	MAX7 	= 	1e+7 + 7;
const ll 	MAXN 	= 	MAX7;
const ll   	INF   	=  	0x7f7f7f7f7f7f7f7f;
const int  	INFi 	=  	0x7f7f7f7f;
const ll 	MOD		=  	1e+9 + 7;

/*vll adj[MAXN];ll visit[MAXN]={};
int dx8[]={0,1,1,1,0,-1,-1,-1}, dy8[]={1,1,0,-1,-1,-1,0,1};
int dx4[]={0,1,0,-1}, dy4[]={1,0,-1,0};
*/
// <------------- Declare Variables Here ------------> //

ll T = 1;
ll n, x, k, m, r;
string s;


// <------------- Implement Functions Here ----------> //



// <------------- Start of main() -------------------> //

void MAIN() {
	cin>> n>> k;
	vll a(n);
	for(ll i=0;i<n;i++) {
		cin >>a[i];
	}
	sort(all(a));
	reverse(all(a));

	ll i=0;
	while(i<n && k) {
		a[i] = 0;
		k--;i++;
	}

	ll sum = 0;
	for(auto it : a) sum += it;
		cout<<sum;
}

int main() {
	fastio; randomINT;
	//cin >> T;
	while (T--) {
		MAIN();
	}
	return 0;
}