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// C++ program to do range minimum query
// using sparse table
#include <bits/stdc++.h>
using namespace std;
#define MAX 100005
// lookup[i][j] is going to store minimum
// value in arr[i..j]. Ideally lookup table
// size should not be fixed and should be
// determined using n Log n. It is kept
// constant to keep code simple.
int lookup[100005][20];
int arr[MAX];
// Fills lookup array lookup[][] in bottom up manner.
void buildSparseTable(int n)
{
// Initialize M for the intervals with length 1
for (int i = 0; i < n; i++)
lookup[i][0] = arr[i];
// Compute values from smaller to bigger intervals
for (int j = 1; (1 << j) <= n; j++) {
// Compute minimum value for all intervals with
// size 2^j
for (int i = 0; (i + (1 << j) - 1) < n; i++) {
// For arr[2][10], we compare arr[lookup[0][7]]
// and arr[lookup[3][10]]
if (lookup[i][j - 1] <
lookup[i + (1 << (j - 1))][j - 1])
lookup[i][j] = lookup[i][j - 1];
else
lookup[i][j] =
lookup[i + (1 << (j - 1))][j - 1];
}
}
}
// Returns minimum of arr[L..R]
int query(int L, int R)
{
// Find highest power of 2 that is smaller
// than or equal to count of elements in given
// range. For [2, 10], j = 3
int j = (int)log2(R - L + 1);
// Compute minimum of last 2^j elements with first
// 2^j elements in range.
// For [2, 10], we compare arr[lookup[0][3]] and
// arr[lookup[3][3]],
if (lookup[L][j] <= lookup[R - (1 << j) + 1][j])
return lookup[L][j];
else
return lookup[R - (1 << j) + 1][j];
}
// Driver program
int main()
{
ifstream fin("rmq.in");
ofstream fout("rmq.out");
#define cin fin
#define cout fout
int n,m;
cin>>n>>m;
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
buildSparseTable(n);
for(int i=0;i<m;i++)
{
int a,b;
cin>>a>>b;
cout<<query(a-1,b-1)<<endl;
}
return 0;
}