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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
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;
}

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);
//const ll mod = 1000000007;
ll N, M, K;

ll imos[1005000];

int main() {
    cin >> N >> M >> K;
    vector<ll> A(N);
    for(int i = 0; i < N; i++) cin >> A[i];
    sort(A.begin(), A.end());
    ll sum = 0;
    for(int i = N - 1; i >= 0; i--) {
        ll rest = M - sum;
        if(rest < A[i]) {
            imos[A[i]]++;
            ll tmp = A[i] - rest;
            imos[A[i]-tmp]--;
            //cerr << i << " " << A[i] << " " << A[i]- tmp - 1 << endl;
        }
        sum += A[i];
        chmin(sum, M);
    }
    for(int i = 1e6; i >= 0; i--) {
        imos[i] += imos[i+1];
    }
    /*
    for(int i = 0; i <= 10; i++) {
        cerr << i << " "<< imos[i] << endl;
    }
    */
    ll ans = 0;
    for(int i = 1e6; i >= 1; i--) {
        ll tmp = min(imos[i], K);
        imos[i] -= tmp;
        K -= tmp;
        //if(i <= 10) cerr << i << " " << imos[i] << endl;
        ans += i * imos[i];
    }
    cout << ans << endl;
    return 0;
}