#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N];
int f[N][16];
int n;
void rmq_init() //建立:dp(i,j) = min{dp(i,j-1),dp(i+2^(j-1),j-1) O(nlogn)
{
for (int i = 1; i <= n; i++)
f[i][0] = a[i];
int k = floor(log((double)n) / log(2.0)); //C/C++取整函数ceil()大,floor()小
for (int j = 1; j <= k; j++)
for (int i = 1; i <= n; i++)
{
if (i + (1 << (j - 1)) <= n) //f(i,j) = min{f(i,j-1),f(i+2^(j-1),j-1)
f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]); ///改为 min 可求最小值
}
}
int rmq(int i, int j) //查询：返回区间[i,j]的最小值
{
int k = floor(log((double)(j - i + 1)) / log(2.0));
return max(f[i][k], f[j - (1 << k) + 1][k]); ///改为 min 可求最小值
}
int main()
{
int icase = 0;
while(scanf("%d", &n)!=EOF)
{
memset(a,0,sizeof(a));
memset(f,0,sizeof(f));
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
rmq_init();
int m;
printf("Case %d:\n",++icase);
scanf("%d",&m);
for(int i = 0; i < m; i++)
{
int x,y;
scanf("%d%d",&x,&y);
printf("%d\n",rmq(x,y));
}
}
}
