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 <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <climits>
#include <cmath>
#include <complex>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <iomanip>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#include <cstdint>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
#define MP make_pair
#define PB push_back
#define inf 1000000007
#define rep(i,n) for(int i = 0; i < (int)(n); ++i)
#define all(x) (x).begin(),(x).end()

template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}
 
template<class T> inline bool chmax(T &a, T b){
    if(a<b){
        a = b;
        return true;
    }
    return false;
}

template<class T> inline bool chmin(T &a, T b){
    if(a>b){
        a = b;
        return true;
    }
    return false;
}

ll dp[1000010];
int main(){
    int n;
    cin >> n;
    ll m,k;
    cin >> m >> k;
    vector<ll>a(n);
    rep(i,n)cin >> a[i];
    sort(a.rbegin(),a.rend());
    bool flag = 1;
    ll sm = 0;
    
    for(int i=0;i<n;i++){
        if(flag){
            if(sm + a[i]<=m){
                sm += a[i];
            }else{
                flag = 0;
                dp[a[i]]++;
                dp[m-sm]--;
            }
        }else{
            dp[a[i]]++;
            dp[0]--;
        }
    }
    ll cnt = 0;
    ll res = 0;
    for(ll i=1000001;i>=0;i--){
        cnt += dp[i];
        if(cnt==0)continue;
        if(k>0){
            if(k>cnt){
                k-=cnt;
            }else{
                res += (cnt-k)*i;
                k = 0;
            }
        }else{
            res += cnt*i;
        }
    }
    cout << res << endl;
    return 0;
}