Problem: Problem statement

The spellbook you have contains $ N $ of spells.

Magic is numbered from $ 1 $ to $ N $, and the cost of magic $ i (1 \ le i \ le N) $ is initially an integer $ A_i $.

Your goal is to cast all the spells in the spellbook $ 1 $ each.

You can eat $ K $ of cookies before you start casting magic. It doesn't take long to eat cookies.

For every $ 1 $ you eat a cookie, you can choose a magic with a positive cost of $ 1 $ and reduce that cost by $ 1 $.

After eating the cookie, you start casting magic.

Your MP is initially $ M $. You repeat either of the following to cast $ N $ of magic $ 1 $ in any order.

* Choose an integer $ i (1 \ le i \ le N) $ for $ 1 $ and cast the magical $ i $. However, the current MP must cost more than the magical $ i $.
* Time does not elapse.
* Consume MP for the cost of magical $ i $.
* Rest. However, if the current MP is $ m $, then $ m <M $ must be.
* Time elapses $ M --m $.
* Recover $ 1 $ MP.



Find the minimum amount of time it will take you to cast $ N $ of spells $ 1 $ each in any order.

Constraint

* $ 1 \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq M \ leq 10 ^ 6 $
* $ 0 \ leq K \ leq \ sum_ {i = 1} ^ {N} A_i $
* $ 1 \ leq A_i \ leq M $
* All inputs are integers.



* * *

input

Input is given from standard input in the following format.


$ N $ $ M $ $ K $
$ A_1 $ $ A_2 $ $ \ ldots $ $ A_N $


output

Print the minimum amount of time it takes to cast $ N $ of magic $ 1 $ each.

* * *

Input example 1


2 4 0
twenty four


Output example 1


3


Since the cost of any magic cannot be reduced, I will continue to cast magic as it is.

* First, cast the magic $ 1 $. It consumes $ 2 $ of MP, so the remaining MP is $ 4-2 = 2 $.



You cannot cast magic $ 2 $ as it is because you need $ 4 $ MP to cast magic $ 2 $.

* I'll take a rest. Time elapses $ 4-2 = 2 $ seconds. It recovers $ 1 $ of MP, so the remaining MP is $ 2 + 1 = 3 $.
* I'll take a rest. Time elapses $ 4-3 = 1 $ seconds. It recovers $ 1 $ of MP, so the remaining MP is $ 3 + 1 = 4 $.
* Cast magic $ 2 $. It consumes $ 4 $ of MP, so the remaining MP is $ 4-4 = 0 $.



From the above, if the time is $ 2 + 1 = 3 $ seconds, the magic $ 1,2 $ can be cast $ 1 $ each time. You can't cast all the spells in less time, so the answer you're looking for is $ 3 $.

* * *

Input example 2


3 9 6
2 3 9


Output example 2


0


With a final magic cost of $ 2, 2, 4 $, you can cast all the magic without a break.

* * *

Input example 3


3 16 2
6 9 9


Output example 3


twenty one


* * *

Input example 4


2 1000000 0
1000000 1000000


Output example 4


500000500000


The answer may not fit in the range that can be represented by a 32-bit integer.





Example

Input

2 4 0
2 4


Output

3

Solution (in CPP):

Solution:
#include <string>
#include <vector>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<stack>
#include<queue>
#include<cmath>
#include<algorithm>
#include<functional>
#include<list>
#include<deque>
#include<bitset>
#include<set>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<cstring>
#include<sstream>
#include<complex>
#include<iomanip>
#include<numeric>
#include<cassert>
#define X first
#define Y second
#define pb push_back
#define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X))
#define reps(X,S,Y) for (int (X) = S;(X) < (Y);++(X))
#define rrep(X,Y) for (int (X) = (Y)-1;(X) >=0;--(X))
#define rreps(X,S,Y) for (int (X) = (Y)-1;(X) >= (S);--(X))
#define repe(X,Y) for ((X) = 0;(X) < (Y);++(X))
#define peat(X,Y) for (;(X) < (Y);++(X))
#define all(X) (X).begin(),(X).end()
#define rall(X) (X).rbegin(),(X).rend()
#define eb emplace_back
#define UNIQUE(X) (X).erase(unique(all(X)),(X).end())
#define Endl endl
#define NL <<"\n"

using namespace std;
using ll=long long;
using pii=pair<int,int>;
using pll=pair<ll,ll>;
template<class T> using vv=vector<vector<T>>;
template<class T> inline bool MX(T &l,const T &r){return l<r?l=r,1:0;}
template<class T> inline bool MN(T &l,const T &r){return l>r?l=r,1:0;}
//#undef NUIP
#ifdef NUIP
#include "benri.h"
#else
#define out(args...)
#endif
#ifdef __cpp_init_captures
template<typename T>vector<T> table(int n, T v){ return vector<T>(n, v);}
template <class... Args> auto table(int n, Args... args){auto val = table(args...); return vector<decltype(val)>(n, move(val));}
#endif
const ll MOD=1e9+7; //998244353

ll sumf(const vector<ll> &a,ll lb){
	ll re=0;
	for(auto x:a) re+=max(0ll,x-lb);
	return re;
}

int main(){
  ios_base::sync_with_stdio(false); cin.tie(0);
  cout<<fixed<<setprecision(0);
	int n,m;
	ll t;
	cin>>n>>m>>t;
	vector<ll> a(n);
	for(auto &x:a) cin>>x;
	sort(all(a));
	int rem=m;
	while(a.size() && rem>=a.back()){
		rem-=a.back();
		a.pop_back();
	}
	if(a.empty()){
		cout<<0 NL;
		return 0;
	}
	out(a,1);
	//keep x==rem if pos
	if(t>sumf(a,rem+1)){
		assert(a.back()>=rem);
		t-=a.back()-rem;
		rem=0;
		a.pop_back();
	}
	if(a.empty()){
		cout<<0 NL;
		return 0;
	}
	out(a,1);

	if(t>=sumf(a,0)){
		cout<<0 NL;
		return 0;
	}
	int l=0,r=MOD;
	while(r-l>1){
		int m=(l+r)/2;
		(sumf(a,m)<=t?r:l)=m;
	}
	out(a,r,1);
	t-=sumf(a,r);
	for(auto &x:a) MN<ll>(x,r);
	rep(i,t) --a[a.size()-1-i];
	sort(rall(a));
	ll re=0;
	for(ll x:a){
		ll t=min<ll>(x,rem);
		rem-=t;
		re+=x*(x+1)/2-t*(t+1)/2;
	}
	cout<<re NL;
  return 0;
}