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<iostream>
#include<cstdio>
#include<algorithm>
#include<cassert>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define MOD2 998244353
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;
Int rest;

Int cnt(vector<Int> &a, Int m){
    Int ans = 0;
    for(int i = 0;i < a.size() - 1;i++){
        ans += max(0ll, a[i] - m + 1);
    }
    ans += a.back() - max(rest, m-1);
    return ans;
}

int main(){
    Int n, m, k;
    cin >> n >> m >> k;
    vector<Int> a(n);
    for(int i = 0;i < n;i++){
        cin >> a[i];
    }
    sort(a.begin(), a.end());

    rest = m;
    while(!a.empty() && a.back() <=rest){
        rest -= a.back();
        a.pop_back();
    }
   
    Int sum = 0;
    for(auto x:a)sum += x;
    if(sum - k - rest <= 0){
        cout << 0 << endl;
        return 0;             
    }
    
    Int bottom = 0, top = a.back() + 1;
    while(top - bottom > 1){
        Int mid = (top + bottom) / 2;
        if(cnt(a, mid) <= k)top = mid;
        else bottom = mid;
    }
    for(int i = 0;i < a.size();i++){
        Int goal = top-1;
        if(i+1 == a.size())goal = max(goal, rest);
        if(a[i] >= goal){
            k -= a[i] - goal;
            a[i] = goal;
        }
    }
    if(a.back() == rest){
        rest = 0;
        a.pop_back();
    }
    for(int i = (int)a.size()-1;i >= 0 && k > 0;i--){
        a[i]--;
        k--;
    }
    Int ans = 0;
    for(auto x:a)ans += (x+1) * x / 2;
    ans -= (rest + 1) * rest / 2;
    cout << ans << endl;
    return 0;
}