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>
#include <iomanip>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
#define ALL(a) a.begin(), a.end()
#define fi first
#define se second
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

ll N,M,H,W,K,Q,A,B;
string S;
const ll MOD = 998244353;
//const ll MOD = (1e+9) + 7;
const ll INF = 1LL<<60;
typedef pair<ll, ll> P;

int main() {
    cin>>N>>M>>K;
    vec a(N);
    rep(i,N) cin>>a[i];
    ll SUM = 0;
    rep(i,N) SUM += a[i];
    if(SUM - K <= M){
        cout<<0<<endl;
        return 0;
    }
    sort(ALL(a));
    reverse(ALL(a));
    ll ans = 0;
    rep(i,N) ans += a[i] * (a[i] + 1) / 2;
    int id = 0, last_sum = 0; //最後に回すやつ（これ含む）
    while(a[id] + last_sum <= M){
        last_sum += a[id];
        ans -= a[id] * (a[id] + 1)/2;
        ++id;
    }
    ll border = M - last_sum;
    ll judge = 0;
    Rreps(i, N, id) if(a[i] > border) judge += a[i] - border;
    if(judge <= K){
        K -= a[id] - border;
        ans -= a[id] * (a[id] + 1) / 2;
        ++id;
    }else{
        ans -= border * (border + 1) / 2;
    }
    a.push_back(-1);
    ll num = 1;
    ll now = a[id];
    //cout<<border<<endl;
    //cout<<ans<<endl;
    Rrep(i, now){
        while(a[id + 1] > i) {
            ++id;
            ++num;
        }
        ans -= min(num, K) * (i + 1);
        K -= num;
        if(K <= 0) break;
    }
    cout<<ans<<endl;
}