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)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return abs(a*b)/gcd(a,b);}
std::vector<bool> IsPrime; void sieve(size_t max){if(max+1 > IsPrime.size())IsPrime.resize(max+1,true);IsPrime[0] = false;IsPrime[1] = false;for(size_t i=2; i*i<=max; ++i)if(IsPrime[i])for(size_t j=2; i*j<=max; ++j)IsPrime[i*j] = false;}
#define roundup(divisor,dividend) (divisor + (dividend - 1)) / dividend
#define all(x) (x).begin(),(x).end()
#define size_t ll
const int mod = 1000000007;
/*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる
vector.unique()は隣接した同じ要素を削除するので先にソートをする

計算量は変わらないが楽できるシリーズ
C++11以降でmin({a,b,c...})で複数個のmic,maxを処理できる
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる

*/
int n,k;
std::vector<ll> h;
int main(){
  cin >> n >> k;
  h.resize(n);
  for (size_t i = 0; i < n; i++) {
    cin >> h[i];
  }
  sort(all(h),std::greater<ll>());
  ll ans=0;
  for (size_t i = k; i < n; i++) {
    ans+=h[i];
  }
  cout << ans << endl;
}