/**
____ ____ ____ ____ ____
||a |||t |||o |||d |||o ||
||__|||__|||__|||__|||__||
|/__\|/__\|/__\|/__\|/__\|
**/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll skim (int i);
void answer (vector <int> v);
void impossible ();
void solve (int N, int K, ll A, int S) {
ll D[N + 1];
int l = 1, r = N + 1;
while (l < r) {
int mid = (l + r) / 2;
D[mid] = skim(mid);
if (D[mid] > A) {
r = mid;
} else {
l = mid + 1;
}
}
for (int i = 1; i <= min(l - 1, K); i++) {
D[i] = skim(i);
}
for (int i = max(1, l - K); i <= min(N, l); i++) {
D[i] = skim(i);
}
if (l < K) {
impossible();
return;
}
if (l <= N) {
ll sum = D[l];
for (int i = 1; i <= K - 1; i++) {
sum += D[i];
}
if (sum <= A * 2) {
vector <int> v(K);
iota(v.begin(), v.end() - 1, 1);
v.back() = l;
answer(v);
return;
}
}
if (l - 1 < K) {
impossible();
return;
}
ll sum = 0;
for (int i = 1; i <= K; i++) {
sum += D[i];
}
vector <int> L(K), R;
iota(L.begin(), L.end(), 1);
for (int i = K; i >= 1 && sum < A; i--) {
sum -= D[L.back()];
L.pop_back();
R.push_back(l - 1 + i - K);
sum += D[R.back()];
}
if (sum < A || 2 * A < sum) {
impossible();
return;
}
vector <int> v = L;
reverse(R.begin(), R.end());
for (int i : R) {
v.push_back(i);
}
answer(v);
}
