Problem Description

N sticks are arranged in a row, and their lengths are a1,a2,...,aN.

There are Q querys. For i-th of them, you can only use sticks between li-th to ri-th. Please output the maximum circumference of all the triangles that you can make with these sticks, or print −1 denoting no triangles you can make.

Input

There are multiple test cases.

Each case starts with a line containing two positive integers N,Q(N,Q≤1e5).

The second line contains N integers, the i-th integer ai(1≤ai≤1e9) of them showing the length of the i-th stick.

Then follow Q lines. i-th of them contains two integers li,ri(1≤li≤ri≤N), meaning that you can only use sticks between li-th to ri-th.

It is guaranteed that the sum of Ns and the sum of Qs in all test cases are both no larger than 4×1e5.

Output

For each test case, output Q lines, each containing an integer denoting the maximum circumference.

Sample Input

5 3
2 5 6 5 2
1 3
2 4
2 5

Sample Output

13
16
16

Solution:

可以发现，无法组成三角形的最劣条件便是斐波那契数列，而在本题数列最多到44项(ai<=1e9)

于是我们便可以用主席树来维护区间第K大，然后暴力询问就OK了

Code:

#include<cstdio>
#include<ctype.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define int long long
using namespace std;
const int N=1e5+1;
int n,q,cnt,a[N],b[N],rt[N];
struct SegTree{
    int tot,ls[N*40],rs[N*40],sz[N*40];
    void clear(){
        tot=0;
        memset(ls,0,sizeof(ls));
        memset(rs,0,sizeof(rs));
        memset(rt,0,sizeof(sz));
    }
    void build(int &q,int l,int r){
        q=++tot;sz[q]=0;
        if(l==r) return ;
        int mid=l+r>>1;
        build(ls[q],l,mid);
        build(rs[q],mid+1,r);
    }
    void ins(int &q,int lst,int l,int r,int x){
        q=++tot;
        ls[q]=ls[lst],rs[q]=rs[lst];
        sz[q]=sz[lst]+1;
        if(l==r) return ;
        int mid=l+r>>1;
        if(mid>=x) ins(ls[q],ls[lst],l,mid,x);
        else ins(rs[q],rs[lst],mid+1,r,x);
    }
    int query(int q,int p,int l,int r,int k){
        if(l==r) return b[l];
        int v=sz[ls[q]]-sz[ls[p]];
        int mid=l+r>>1;
        if(v>=k) return query(ls[q],ls[p],l,mid,k);
        else return query(rs[q],rs[p],mid+1,r,k-v);
    }
}T;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
void solve(){T.clear();
    for(int i=1;i<=n;i++)
        a[i]=read(),b[i]=a[i];
    sort(b+1,b+n+1);
    cnt=unique(b+1,b+n+1)-b-1;
    T.build(rt[0],1,cnt);
    for(int i=1;i<=n;i++){
        int v=lower_bound(b+1,b+cnt+1,a[i])-b;
        T.ins(rt[i],rt[i-1],1,cnt,v);
    }
    for(int i=1;i<=q;i++){
        int l=read(),r=read(),u=r-l+1,flag=0;
        if(u<=2){puts("-1");continue;}
        int a=T.query(rt[r],rt[l-1],1,cnt,u);
        int b=T.query(rt[r],rt[l-1],1,cnt,--u);
        while(u>1){
            int v=T.query(rt[r],rt[l-1],1,cnt,--u);
            if(v+b>a){
                printf("%lld\n",a+b+v);
                flag=1;break;
            }a=b,b=v;
        }if(!flag) puts("-1");
    }
}
signed main(){
    while(~scanf("%lld%lld",&n,&q)) solve();
    return 0;
}