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)

/// Bismillahir Rahmanir Rahim
//Author: Tanvir Hussain
//ICE,NSTU
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
const long long MOD = 1000000007;
#define SET(x) memset(x, 0, sizeof(x))
#define SET2d(x,m,n) memset(x, 0, sizeof(x[0][0]) * m * n)
#define SETBOOL(x) memset(x,false,sizeof(x))
#define CLR(x) memset(x, -1, sizeof(x))
#define mp make_pair
#define PII pair<int, int>
#define pf printf
#define sf scanf
#define ALL(x) x.begin(),x.end()
#define pb push_back
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define np std::string::npos
#define highest(x) numeric_limits<x>::max()
#define lowest(x) numeric_limits<x>::min()
#define Inf INFINITY
#define minv(v) *min_element(v.begin(),v.end())
#define maxv(v) *max_element(v.begin(),v.end())
#define cases(cs,t) for(int cs=1;cs<=t;cs++)
#define PI acos(-1)
#define no1 __builtin_popcount
#define BOUNDARY(i, j) ((i >= 0 && i < row) && (j >= 0 && j < column))
#define uniq(vec) vec.resize(distance(vec.begin(),unique(vec.begin(),vec.end())))
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define sz(a) int(a.size())
#define ff first
#define ss second
#define endl "\n"
#define forch(it,s) for(auto it:s)
#define each(it,s) for(auto it = s.begin(); it != s.end(); ++it)
#define rep(i,a) for(int i=0; i<a;i++)
#define rep1(i,a,b) for(int i=(a);i<=(b);++i)
#define irep(i,b,a) for(int i=(b);i>=(a);--i)
#define bits(n) __builtin_popcount(n)
#define maxpq priority_queue<int>
#define minpq priority_queue<int, vector<int>, greater<int> >


int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a);}
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pair<int,int>> vpii;
typedef tree<pair<int, int>,null_type,less<pair<int, int>>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1,-1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
const int maxx=100005;

//this fuction sorts vector pair according to first element in descending order.
bool sortinrev(const pair<int,int> &a,const pair<int,int> &b)
{
    return a.first>b.first;
}

template<typename T>inline T Bigmod(T base, T power, T MOD){
    T ret=1;
    while(power)
    {
        if(power & 1)ret=(ret*base)%MOD;
        base=(base*base)%MOD;
        power>>=1;
    }
    return ret;
}
double sq(double x) {return x*x;}
void solve(){
    ll n,k;
    cin>>n>>k;
    vll vec(n);
    rep(i,n) cin>>vec[i];
    sort(ALL(vec),greater<int>());
    ll sum=0;
    rep(i,n){
        if(i<k) continue;
        sum+=vec[i];
    }
    cout<<sum<<endl;
}

signed main()

{
    IOS;
    /*#ifndef ONLINE_JUDGE
        freopen ("data.in","r",stdin);
        freopen ("data.out","w",stdout);
    #endif*/
    int t;
    t=1;
    while(t--){
        solve();
    }

     return 0;

}
///Alhamdulillah