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>
#define speedio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;

typedef long long       ll;
typedef long double     ld;

typedef vector<int>     vi;
typedef vector<bool>    vb;
typedef vector<ll>      vll;
typedef vector<vi>      vvi;
typedef vector<vll>     vvll;
typedef vector<string>  vs;

typedef pair<int, int>  pii;
typedef pair<ll, int>   pli;
typedef pair<ll, ll>    pll;
typedef vector< pii >   vpi;
typedef vector< pll >   vpll;

// typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;

#define pb           push_back
#define mod          1000000007
#define INF          1e17
#define mini         INT_MIN
#define maxo         INT_MAX
#define eb           emplace_back
#define ff           first
#define ss           second
#define re           return 
#define endl         "\n" 

#define all(c)       c.begin(),c.end()
#define be           begin()
#define en           end()
#define sz(x)        (ll)x.size()
#define mem(a,val)   memset(a,val,sizeof(a))
#define mk(x,y)      make_pair(x,y)
#define mid(s,e)     ((s)+((e)-(s))/2) 

#define fo(i,a,b)     for(ll i = (ll)(a); i <= (ll)(b); i++)
#define rf(i,a,b)    for(ll i = (ll)(a); i >= (ll)(b); i--)
#define read(a,n)    for(int i=0;i<n;i++)cin>>a[i];
#define show(a,n)    for(int i=0;i<n;i++)cout<<a[i]<<' ';cout<<endl;
#define p2(n)        (1LL << (n))
#define deb(x) cout <<#x << " = " << x << " | " << "LINE: "<<__LINE__<< endl;
template<typename... T> void rd(T&... args) { ((cin >> args), ...); }
template<typename... T> void pp(T... args) { ((cout << args << " "), ...); cout<<"\n"; }

void add(ll &x, ll y){ x += y; if(x>=mod) x %= mod; if(x<0) x = x%mod + mod; }

//============== *** START OF CODE ***




void solve()
{
    ll n,k;rd(n,k);
    ll a[n]; read(a,n);
    sort(a,a+n);
    ll left = n-k;
    ll s = 0;
    fo(i,0,left-1){
        s+=a[i];
    }
    pp(s);
    
}


void prep(){}

//=============== *** END OF CODE ***

int main(){
    speedio; cout<<fixed<<setprecision(12);

// #ifndef ONLINE_JUDGE
//     freopen("input.txt", "r", stdin);
//     freopen("output.txt", "w", stdout);
// #endif

    prep();
    int t = 1;
    // cin >> t;
    while(t--)
        solve();

    cout<<"\n";
    return 0;
}