#include <iostream> 
#include <vector> 
#include <algorithm> 
#include <cmath> 
using namespace std;
typedef long long ll;
bool check_k0(int x, int n, int m, const vector<int> &a) {
	ll operations_needed = 0;
	vector<ll> diff(n + 2, 0);
	ll current = 0;
	for (int j=1; j<=n; ++j) {
		current += diff[j];
		if(current < a[j-1]) {
			ll need = a[j-1] - current;
			operations_needed += need;
			if(operations_needed > m) return false;
			int s = min(j + x, n);
			diff[j] += need;
			diff[s +1] -= need;
			current += need;
		}
	}
	return operations_needed <= m;
}
bool check_k1(int x, int n, int m, const vector<int> &a) {
	ll max_required_x = 0;
	for (int j=1; j<=n; ++j) {
		ll required = ( (ll)a[j-1] + m -1 ) / m + j -1;
		if(x < required) {
			return false;
		}
	}
	return true;
}
bool check_k2(int x, int n, int m, const vector<int> &a){
    vector<ll> diff_c(n+2,0);
    vector<ll> diff_ci(n+2,0);
    vector<ll> diff_ci2(n+2,0);
    ll sum_c =0, sum_ci=0, sum_ci2=0;
    ll operations =0;
    for(int j=1; j<=n; j++){
        sum_c += diff_c[j];
        sum_ci += diff_ci[j];
        sum_ci2 += diff_ci2[j];
        ll b_j = sum_c * (ll)j * j + (-2LL *x * sum_c -2LL * sum_ci) * (ll)j + sum_c * (ll)x *x + 2LL *x * sum_ci + sum_ci2;
        if(b_j < a[j-1]){
            ll deficit = a[j-1] - b_j;
            if(x ==0){
                return false; 
            }
            ll need = (deficit + (ll)x *x -1) / ((ll)x *x);
            operations += need;
            if(operations >m) return false;
            
            diff_c[j] += need;
            if(j +x +1 <=n){
                diff_c[j +x +1] -= need;
            }
            diff_ci[j] += need * j;
            if(j +x +1 <=n){
                diff_ci[j +x +1] -= need * j;
            }
            diff_ci2[j] += need * (ll)j * j;
            if(j +x +1 <=n){
                diff_ci2[j +x +1] -= need * (ll)j * j;
            }
            
            sum_c += need;
            sum_ci += need * j;
            sum_ci2 += need * (ll)j * j;
            
            b_j += need * (ll)x *x - 2LL *x * need *j + need * (ll)j *j;
            
        }
    }
    return operations <=m;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, m, k;
	cin >> n >> m >> k;
	vector<int> a(n);
	for (auto &x : a) cin >> x;
	int left = 0, right = 200000;
	int answer = -1;
	while(left <= right) {
		int mid = left + (right - left)/2;
		bool feasible = false;
		if(k ==0) {
			feasible = check_k0(mid, n, m, a);
		} else if(k ==1) {
			feasible = check_k1(mid, n, m, a);
		} else if(k ==2) {
			feasible = check_k2(mid, n, m, a);
		}
		if(feasible) {
			answer = mid;
			right = mid -1;
		} else {
			left = mid +1;
		}
	}
	cout << answer;
}