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)

#pragma GCC optimize("Ofast","unroll-loops")
#pragma GCC push_options
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
// #include <xmmintrin.h>
// #include <immintrin.h>
using namespace::std;
__attribute__((constructor))void init(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
#include<ext/pb_ds/tag_and_trait.hpp>
// #include <boost/multiprecision/cpp_dec_float.hpp>
// #include <boost/multiprecision/cpp_int.hpp>
// namespace mp = boost::multiprecision;
// typedef mp::number<mp::cpp_dec_float<0>> cdouble;
// typedef mp::cpp_int cint;
template<typename T>using pbds=__gnu_pbds::tree<T,__gnu_pbds::null_type,less<T>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
template<typename T>using pbds_map=__gnu_pbds::tree<T,T,less<T>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
template<typename T,typename E>using hash_map=__gnu_pbds::gp_hash_table<T,E>;
template<typename T>using pqueue =__gnu_pbds::priority_queue<T, less<T>,__gnu_pbds::rc_binomial_heap_tag>;
typedef long long lint;
#define INF (1LL<<60)
#define IINF (1<<30)
#define EPS (1e-10)
#define MOD 1000000007LL
typedef vector<lint> vec;
typedef vector<vector<lint>> mat;
typedef vector<string> svec;
typedef vector<vector<string>> smat;
template<typename T>inline void numout(T t){bool f=0;for(auto i:t){cout<<(f?" ":"")<<i<INF/2?i:"INF";f=1;}cout<<endl;}
template<typename T>inline void numout2(T t){for(auto i:t)numout(i);}
template<typename T>inline void output(T t){bool f=0;for(auto i:t){cout<<(f?" ":"")<<i;f=1;}cout<<endl;}
template<typename T>inline void output2(T t){for(auto i:t)output(i);}
template<typename T>inline void _output(T t){bool f=0;for(lint i=0;i<t.size();i++){cout<<f?"":" "<<t[i];f=1;}cout<<endl;}
template<typename T>inline void _output2(T t){for(lint i=0;i<t.size();i++)output(t[i]);}
#define rep(i,n) for(lint i=0;i<lint(n);++i)
#define repi(i,a,b) for(lint i=lint(a);i<(lint)(b);++i)
#define rrep(i,n) for(lint i=lint(n)-1;i>=0;--i)
#define rrepi(i,a,b) for(lint i=lint(b)-1;i>=lint(a);--i)
#define irep(i) for(lint i=0;;++i)
#define all(n) begin(n),end(n)
#define dist(a,b,c,d) sqrt(pow(a-c,2)+pow(b-d,2))
inline lint gcd(lint A,lint B){return B?gcd(B,A%B):A;}
inline lint lcm(lint A,lint B){return A/gcd(A,B)*B;}
// inline cint cgcd(cint A,cint B){return B?cgcd(B,A%B):A;}
// inline cint clcm(cint A,cint B){return A/cgcd(A,B)*B;}
inline bool chmin(auto& s,const auto& t){bool res=s>t;s=min(s,t);return res;}
inline bool chmax(auto& s,const auto& t){bool res=s<t;s=max(s,t);return res;}
const vector<lint> dx={1,0,-1,0,1,1,-1,-1};
const vector<lint> dy={0,1,0,-1,1,-1,1,-1};
#define SUM(v) accumulate(all(v),0LL)
auto call=[&](auto f,auto... args){return f(f,args...);};

int main(){
    lint n,k;
    cin>>n>>k;
    vec a(n);
    rep(i,n)cin>>a[i];
    sort(all(a));
    reverse(all(a));
    lint ans=0;
    rep(i,n){
        if(i>=k)ans+=a[i];
    }
    cout<<ans<<endl;
}