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# 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 ;
}