#include <stdio.h>
#include <vector>
 
using namespace std;
 
#define NMAX 100002
#define INF 0x3f3f3f3f
 
long int M[NMAX*3];
long int A[NMAX];
long int N,m,mx;
 

 void initialize(long int node, int b, int e)
  {
      if (b == e)
          M[node] = b;
      else
       {
  //compute the values in the left and right subtrees
          initialize(2 * node, b, (b + e) / 2);
          initialize(2 * node + 1, (b + e) / 2 + 1, e);
  //search for the minimum value in the first and 
  //second half of the interval
          if (A[M[2 * node]] <= A[M[2 * node + 1]])
              M[node] = M[2 * node];
          else
              M[node] = M[2 * node + 1]; 
      }
  } 


int query(int node, int b, int e, int i, int j)
  {
      int p1, p2;
   
  //if the current interval doesn't intersect 
  //the query interval return -1
      if (i > e || j < b)
          return -1;
   
  //if the current interval is included in 
  //the query interval return M[node]
      if (b >= i && e <= j)
          return M[node];
   
  //compute the minimum position in the 
  //left and right part of the interval
      p1 = query(2 * node, b, (b + e) / 2, i, j);
      p2 = query(2 * node + 1, (b + e) / 2 + 1, e, i, j);
   
  //return the position where the overall 
  //minimum is
      if (p1 == -1)
          return M[node] = p2;
      if (p2 == -1)
          return M[node] = p1;
      if (A[p1] <= A[p2])
          return M[node] = p1;
      return M[node] = p2;
  }


 
int main()
{
    freopen("rmq.in","r",stdin);
    freopen("rmq.out","w",stdout);
 
    long int i,x,y,k;
 
    scanf("%ld %ld",&N,&m);
    for (i=1;i<=N;i++)
        {
            scanf("%ld",&x);
            A[i]=x;
        }
 	initialize(1,1,N);
    for (i=1;i<=m;i++)
        {
            scanf("%ld %ld",&x,&y);
             
            mx = A[query(1,1,N,x,y)];
            printf("%ld\n",mx);
                 
        }
    return 0;       
}