#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int a[N];
int n, q;

/*ST跳表：快速查询区间内的极大极小值*/
int st1[N][40], st2[N][40];
void build() {
    for (int i = 1; i <= n; i++) st1[i][0] = a[i], st2[i][0] = a[i];
    for (int j = 1; (1 << j) <= n; j++)
        for (int i = 1; (i + (1 << j) - 1) <= n; i++) {
            st1[i][j] = max(st1[i][j - 1], st1[i + (1 << (j - 1))][j - 1]);
            st2[i][j] = min(st2[i][j - 1], st2[i + (1 << (j - 1))][j - 1]);
        }
}
int queryMax(int x, int y) {
    int t = log2(y - x + 1);
    return max(st1[x][t], st1[y - (1 << t) + 1][t]);
}
int queryMin(int x, int y) {
    int t = log2(y - x + 1);
    return min(st2[x][t], st2[y - (1 << t) + 1][t]);
}

/*
5
50 60 45 39 78
2 5

答案应该是45
*/

signed main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    build();
    int x, y;
    scanf("%d %d", &x, &y);
    cout << queryMax(x, y) << endl;
    cout << queryMin(x, y) << endl;
}