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 <utility>
#include <tuple>
#include <vector>
#include <string>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <algorithm>
#include <functional>
#include <climits>
#include <numeric>
#include <queue>
#include <cmath>
#include <iomanip>
#include <array>
#include <string>
#include <stack>
#include <cassert>
#include <memory>
long long int recover_all(int a) {
	return a * (a + 1LL) / 2;
}
int main() {
	int n, m; long long int k; std::cin >> n >> m >> k;
	std::vector<int> magic(n); for (auto& a : magic) std::cin >> a;
	std::sort(magic.begin(), magic.end());
	const auto need_to_recover = std::accumulate(magic.begin(), magic.end(), 0LL) - m;
	if (need_to_recover <= 0) {
		std::cout << 0 << '\n'; return 0;
	}
	int min{ 0 }, max{ m };
	while (min < max) {
		const auto mid = (min + max) >> 1;
		auto cookie = k;
		auto recover = need_to_recover;
		for (auto a : magic) {
			if (cookie < 0) break;
			if (mid < a) {
				const auto minus = std::min<long long int>(a - mid, recover);
				cookie -= minus;
				recover -= minus;
				a -= minus;
			}
			recover -= std::min<long long int>(recover, a);
		}
		if (cookie < 0) {
			min = mid + 1;
		}
		else {
			max = mid;
		}
	}
	auto cookie = k;
	auto recover = need_to_recover;
	long long int time{ 0 };
	int just{ 0 };
	for (auto a : magic) {
		if (max < a) {
			const auto minus = std::min<long long int>(a - max, recover);
			cookie -= minus;
			recover -= minus;
			a -= minus;
		}
		const auto r = std::min<long long int>(recover, a);
		time += recover_all(a) - recover_all(a - r);
		recover -= r;
		if (a == max) ++just;
	}
	const auto additional = std::min<long long int>(just, cookie);
	time -= max * additional;
	std::cout << time << '\n';
}