#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=3e5+10,mod=1e6+3;
int a[N],b[N],id[N],siz,n,q;
int se(int u){return lower_bound(b+1,b+siz+1,u)-b;};
struct po{int l,r,sum;}tr[N<<5];
int idx=0,root[N<<5];
// 难道没有极端样例吗,就是每次查询1-n,且1-n都不满足;
void update(int &x,int y,int l,int r,int t)
{
x=++idx;
tr[idx]=tr[y];
tr[idx].sum++;
if(l==r)return ;
int mid=l+r>>1;
if(t<=mid)update(tr[idx].l,tr[y].l,l,mid,t);
else update(tr[idx].r,tr[y].r,mid+1,r,t);
}
int quary(int u,int v,int l,int r,int t)
{
int sum=tr[tr[u].l].sum-tr[tr[v].l].sum;
if(l==r)return l;
int mid=l+r>>1;
if(t<=sum)return quary(tr[u].l,tr[v].l,l,mid,t);
else quary(tr[u].r,tr[v].r,mid+1,r,t-sum);
}
int main()
{
while(~scanf("%d%d",&n,&q))
{
idx=0;
for(int i=1;i<=n;i++)scanf("%d",&a[i]),b[i]=a[i],root[i]=0;
sort(b+1,b+n+1);
siz=unique(b+1,b+n+1)-b-1;
for(int i=1;i<=n;i++)id[i]=se(a[i]);
for(int i=1;i<=n;i++)
update(root[i],root[i-1],1,siz,id[i]);// 主席树;
while(q--)
{
int l,r;
scanf("%d%d",&r,&l);
bool flag=0;
if(l-r+1<3){printf("-1\n");continue;}
int x=quary(root[l],root[r-1],1,siz,l-r+1),y=quary(root[l],root[r-1],1,siz,l-r),z;
for(int i=l-2;i>=r;i--)
{
z=quary(root[l],root[r-1],1,siz,i-r+1);// 第i大的元素;
if(b[z]+b[y]>b[x]){
flag=1;
break;
}
x=y,y=z;
}
printf("%lld\n",flag?(ll)b[x]+(ll)b[y]+(ll)b[z]:-1);// 爆int
}
}
return 0;
}