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#include <bits/stdc++.h>
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using namespace std;
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const int N = 100010;
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int a[N];
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int n, q;
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/*ST跳表：快速查询区间内的极大极小值*/
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int st1[N][40], st2[N][40];
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void build() {
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for (int i = 1; i <= n; i++) st1[i][0] = a[i], st2[i][0] = a[i];
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for (int j = 1; (1 << j) <= n; j++)
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for (int i = 1; (i + (1 << j) - 1) <= n; i++) {
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st1[i][j] = max(st1[i][j - 1], st1[i + (1 << (j - 1))][j - 1]);
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st2[i][j] = min(st2[i][j - 1], st2[i + (1 << (j - 1))][j - 1]);
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}
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}
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int queryMax(int x, int y) {
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int t = log2(y - x + 1);
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return max(st1[x][t], st1[y - (1 << t) + 1][t]);
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}
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int queryMin(int x, int y) {
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int t = log2(y - x + 1);
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return min(st2[x][t], st2[y - (1 << t) + 1][t]);
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}
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/*
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5
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50 60 45 39 78
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2 5
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答案应该是45
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*/
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signed main() {
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scanf("%d", &n);
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for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
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build();
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int x, y;
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scanf("%d %d", &x, &y);
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cout << queryMax(x, y) << endl;
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cout << queryMin(x, y) << endl;
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}
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