You are given a list of numbers A₁ A₂ .. A_N and M queries. For the i-th query:

• The query has two parameters L_i and R_i.
• The query will define a function F_i(x) on the domain [L_i, R_i] ∈ Z.
• F_i(L_i) = A_Li
• F_i(L_i + 1) = A_{(Li + 1)}
• for all x >= L_i + 2, F_i(x) = F_i(x - 1) + F_i(x - 2) × A_x

You task is to calculate F_i(R_i) for each query. Because the answer can be very large, you should output the remainder of the answer divided by 1000000007.

1
4 7
1 2 3 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4

1
2
5
13
11
4
4

#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<algorithm>

#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
using namespace std;
typedef long long ll;

const int mod=1e9+7;
const int maxn=1e5+10;
int n,m;
/*
忘了线段树这个优美的结构了，这道题，当时真是可惜，在自己的能力范围内，但是没有做出来
当时，我想了一下线段树，但是很快就又跳了过去，主要是考虑到复杂度，不知道当时自己怎么想的
感觉就是会超，它是一个log(n)的复杂度，是没有问题的 
*/

struct Martix{
	ll mar[3][3];
	Martix operator *(const Martix &b){
		Martix c;
		for(int i=1;i<=2;i++){
			for(int j=1;j<=2;j++){
				c.mar[i][j]=0;
				for(int k=1;k<=2;k++){
					c.mar[i][j]=(c.mar[i][j]+mar[i][k]*b.mar[k][j]%mod)%mod;
				}
			}
		}
		return c;
	} 
};
struct node{
	Martix m;
	int l,r;
}tr[maxn*4];

ll val[maxn];

void build(int l,int r,int o){
	tr[o].l=l,tr[o].r=r;
	if(l==r){
		tr[o].m.mar[1][1]=1;
		tr[o].m.mar[2][1]=1;
		tr[o].m.mar[1][2]=val[n-l+1];
		tr[o].m.mar[2][2]=0;
		return ;
	} 
	int mid=(l+r)>>1;
	build(l,mid,ls(o));
	build(mid+1,r,rs(o));
	tr[o].m=tr[ls(o)].m*tr[rs(o)].m;
}

Martix query(int l,int r,int o){
	if(tr[o].l>=l&&tr[o].r<=r)
		return tr[o].m;
	int mid=(tr[o].l+tr[o].r)>>1;
	if(r<=mid)
		return query(l,r,ls(o));
	else if(l>mid)
		return query(l,r,rs(o));
	else
		return query(l,mid,ls(o))*query(mid+1,r,rs(o));
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
    	scanf("%d %d",&n,&m);
    	for(int i=1;i<=n;i++){
    		scanf("%lld",&val[i]);
    	}
    	build(1,n,1);
    	for(int i=1;i<=m;i++){
    		int l,r;
    		scanf("%d %d",&l,&r);
			if(r-l<2){
				printf("%lld\n",val[r]);
			} 
			else{
				Martix m=query(n-r+1,n-l-1,1);
				ll ans=(val[l+1]*m.mar[1][1]+val[l]*m.mar[1][2]%mod+mod)%mod;
				printf("%lld\n",ans);
			}
    	}
    }
    return 0;
}