#pragma GCC optimize("Ofast")
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <bitset>
#include <cmath>
#include <queue>
#include <ctime>
#include <set>
#include <map>
#define int long long
#define rep(i, a, b) for (int i = (a); i <= (b); i ++ )
#define rop(i, a, b) for (int i = (a); i < (b); i ++ )
#define dep(i, a, b) for (int i = (a); i >= (b); i -- )
#define dop(i, a, b) for (int i = (a); i > (b); i -- )

#define Darr(a, L, R) cerr << #a "[" << L << " ~ " << R << "] = "; rep(x, L, R) cerr << a[x] << " "; cerr << '\n';
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define D(x) (cerr << #x << " = " << x << '\n')
#define vit vector<int>::iterator
#define all(x) x.begin(), x.end()
#define min(a, b) (a < b ? a : b)
#define max(a, b) (a > b ? a : b)
#define chkmin(a, b) (a = min(a, b))
#define chkmax(a, b) (a = max(a, b))
#define sit set<int>::iterator
#define lowbit(x) (x & -x)
#define eb emplace_back
#define ub upper_bound
#define pb push_back
#define pi acos(-1)
#define gc getchar
#define pc putchar
#define db double
#define y second
#define x first

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<db, db> PDD;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
const int mod = 998244353;
const int eps = 1e-9;

const int N = 1000010; int T = 1;
void print(int x) { dep(i, 20, 0) pc(((x >> i) & 1) ? '1' : '0'); }
int qpow(int a, int b = mod - 2, int s = 1) { for (; b; b >>= 1, a = 1ll * a * a % mod) if (b & 1) s = 1ll * s * a % mod; return s; }
namespace IO {
	void read() { return; }
	void write(char ch) { pc(ch); return; }
	void write() { return; }
	template <typename T> void read(T &x) { x = 0; T w = 0; char ch = gc(); while (ch < '0' || ch > '9') w |= (ch == '-'), ch = gc(); while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = gc(); x = w ? -x : x; }
	template <typename T> void print(T x) { if (!x) return; print<T>(x / 10), pc((x % 10) ^ '0'); }
	template <typename T> void write(T x) { if (x > 0) print<T>(x); else if (x < 0) pc('-'), print<T>(-x); else pc('0'); }
	template <typename T> void write(T x, char en) { write<T>(x), pc(en); }
	template <typename T, typename ...T2> void read(T &s, T2 &...oth) { read(s); read(oth...); return; }
}; using namespace IO;

const int INF = 1e15;
int lb[N], rb[N], a[N], stk[N];
int n, L, R, top, a1, a2;
vector<int> ans;
int calc(int i, int lim) {
	// 以 i 为中心长度 <= lim 的方案数 
	int r = rb[i], l = lb[i];
	if (lim == 0) return 0;
	if (r - l + 1 <= lim) return (i - l + 1) * (r - i + 1);
	lim ++ ; int r2 = min(r, i + lim - 1);
	int s = (i - l + 1) * (r - r2); int u = (r2 - lim + 1 - l + 1);
	int v = (i - l + 1 <= lim) ? 1 : i - lim + 1 - l + 1;
	s = s + (u + v) * (u - v + 1) / 2; 
	return (i - l + 1) * (r - i + 1) - s;
}
int Rank(int mid) {
	// 面积为 mid 的矩形排名
	int s = 0; rep(i, 1, n)
		s += calc(i, (mid + a[i] - 1) / a[i] - 1);
	return s + 1;
}
void sub_intake(int i, int ll, int lr) {
	// 将 i 为中心，长度在 [ll, lr] 中的矩形统计进答案。
	if (ll > lr) return;
	int l = lb[i], r = rb[i], len = r - l + 1;
	if (len < ll) return;
	l = max(l, i - lr + 1);
	rep(it, l, i) {
		int tl = it + ll - 1, tr = it + lr - 1;
		tr = min(tr, r); tl = max(tl, i); if (tl > tr) break;
		rep(j, tl, tr) ans.pb((j - it + 1) * a[i]);
	}
}
void intake(int sl, int sr) {
	// 将面积在 [sl, sr] 中的矩形统计进答案。
	rep(i, 1, n) sub_intake(i, (sl + a[i] - 1) / a[i], sr / a[i]);
}
void sub() {
	read(n); 
	rep(i, 1, n) read(a[i]); 
	read(L, R);
	// 第一遍单调栈，处理每个位置左边第一个小于**等于**这个数的位置。
	top = 0; rep(i, 0, n) {
		while (top and a[stk[top]] > a[i]) top -- ;
		lb[i] = stk[top] + 1; stk[ ++ top] = i;
	}
	// 第二遍单调栈，处理每个位置右边第一个小于这个数的位置。
	top = 0; dep(i, n + 1, 1) {
		while (top and a[stk[top]] >= a[i]) top -- ;
		rb[i] = stk[top] - 1; stk[ ++ top] = i;
	} 
	// 找到面积大于 area(L) 的最小的矩形面积 
	int l = 0, r = (int)1e13;
	while (l < r) {
		int mid = l + r >> 1;
		if (Rank(mid) > L) r = mid;
		else l = mid + 1;
	} a1 = l;
	// 找到面积小于 area(R) 的最大矩形面积
	l = 0, r = (int)1e13;
	while (l < r) {
		int mid = l + r + 1 >> 1;
		if (Rank(mid + 1) <= R) l = mid;
		else r = mid - 1;
	} a2 = l; 
	intake(a1, a2);
	int area1 = -INF, area2 = INF;
	rep(i, 1, n) {
		int len = (a1 + a[i] - 1) / a[i] - 1;
		if (rb[i] - lb[i] + 1 >= len) area1 = max(area1, len * a[i]);
		len = (a2 / a[i]) + 1;
		if (rb[i] - lb[i] + 1 >= len) area2 = min(area2, len * a[i]);
	} int rk1 = Rank(a1), rk2 = Rank(a2 + 1);
	if (rk1 > R) {
		rep(i, L, R) write(area1, ' ');
		puts(""); return;
	}
	rop(i, L, rk1) ans.pb(area1);
	rep(i, rk2, R) ans.pb(area2);
	sort(all(ans)); for (auto i : ans) printf("%lld ", i);
	return;
}
signed main() {
	while (T -- ) sub();
	return 0;
}