#include<stdio.h>
#include<math.h>

#define nmax 100010
#define nlog 32

int n,m,M[nmax][nlog],a[nmax];
double l2=log(2);

void cit();
void init();
void solve();
int rmq(int,int,int);
int min(int,int);

int main()
{
    cit();
    
    init();
    
    solve();
    
    return 0;
}

int rmq(int i,int j,int k)
{
   // printf("%d %d %d\n",i,j,k);
    return ( min ( M[i][k] , M[j-(1<<k)+1][k]) );
}

void solve()
{
    int k,b,c;
    for(int i=1;i<=m;++i)
        {
        scanf("%d%d",&b,&c);
        k=(int)(log(c-b+1)/l2);
        printf("%d\n",rmq( b , c , k ));
        }
}

void init()
{
    for(int i=1 ; (1<<i)<=n ; ++i )
        for(int j=1;j<=n;++j)
            M[j][i]=rmq( j , min( j+(1<<i) -1 , n ) , (i-1) );
            
  /*  for(int i=0; i<=3;++i)
        {
        for(int j=1;j<=n;++j)
            printf("%d ",M[j][i]);
        printf("\n");
        }*/
}

void cit()
{
    freopen("rmq.in","r",stdin);
    freopen("rmq.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i)
        {
        scanf("%d",&a[i]);
        M[i][0]=a[i];
        }
}

int min(int a,int b)
{
    if (a<b)
        return a;
    return b;
}