#include<bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1000000007;
int B[200005],C[200005],F[500005],F1[500005],F2[500005];
struct node{
	int l,r;
	int a2,b2,c2,ab,ac,bc;
	int kb,kc;
	void addB(int k){
		a2 = (a2+k*k%mod*b2+2*k*ab)%mod;
		ab = (ab+k*b2)%mod;
		ac = (ac+k*bc)%mod;
		kb = (kb+k)%mod;
	}void addC(int k){
		a2 = (a2+k*k%mod*c2+2*k*ac)%mod;
		ac = (ac+k*c2)%mod;
		ab = (ab+k*bc)%mod;
		kc = (kc+k)%mod;
	}
}tree[1000005];
void push_up(int k){
	tree[k].a2 = (tree[k<<1].a2+tree[k<<1|1].a2)%mod;
	tree[k].ab= (tree[k<<1].ab+tree[k<<1|1].ab)%mod;
	tree[k].ac = (tree[k<<1].ac+tree[k<<1|1].ac)%mod; 
}void push_down(int k){
	if(tree[k].kb){
		tree[k<<1].addB(tree[k].kb);
		tree[k<<1|1].addB(tree[k].kb);
		tree[k].kb = 0;
	}if(tree[k].kc){
		tree[k<<1].addC(tree[k].kc);
		tree[k<<1|1].addC(tree[k].kc);
		tree[k].kc = 0;
	}
}void build(int l,int r,int k){
	tree[k].l= l ,tree[k].r = r;
	if(l+1 == r){
		tree[k].b2 = B[l]*B[l]%mod;
		tree[k].c2 = C[l]*C[l]%mod;
		tree[k].bc = B[l]*C[l]%mod;
		return;
	}build(l,l+r>>1,k<<1),build(l+r>>1,r,k<<1|1);
	tree[k].b2 = (tree[k<<1].b2+tree[k<<1|1].b2)%mod;
	tree[k].c2 = (tree[k<<1].c2+tree[k<<1|1].c2)%mod;
	tree[k].bc = (tree[k<<1].bc+tree[k<<1|1].bc)%mod; 
}void update(int l,int r,int k,int kb,int kc){
	int ll = tree[k].l,rr = tree[k].r;
	if(ll>=r or l>=rr)return;
	if(l<=ll and rr<=r){
		tree[k].addB(kb);
		//cout <<" " << k << " " << l << " " << r << " " << tree[k].a2  << " " << tree[k].ab << " " << tree[k].ac << endl;
		tree[k].addC(kc);
		//cout <<" " << k << " " << l << " " << r << " " << tree[k].a2  << " " << tree[k].ab << " " << tree[k].ac << endl;
		return;
	}push_down(k);
	update(l,r,k<<1,kb,kc);update(l,r,k<<1|1,kb,kc);
	push_up(k); 
}int query(int l,int r,int k){
	int ll = tree[k].l,rr = tree[k].r;
	if(ll>=r or l>=rr)return 0;
	if(l<=ll and rr<=r)return tree[k].a2;
	push_down(k);
	return (query(l,r,k<<1)+query(l,r,k<<1|1))%mod;
}int n,q;//int aa[5005];
signed main(){
	srand(time(0));
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//	n = 50,q = 500000;
	cin>> n >> q;
	F[1] = F[2] = 1;
	F1[1] = 1,F2[2] = 1;
	for(int i = 3;i<=500000;i++)F[i] = (F[i-1]+F[i-2])%mod;
	for(int i = 3;i<=500000;i++){
		F1[i] = (F1[i-2]-F1[i-1]+mod)%mod;
		F2[i] = (F2[i-2]-F2[i-1]+mod)%mod;
	}for(int i = 1;i<=n;i++)B[i] = F[i],C[i] = F[i-1];
	build(1,n+1,1);
	while(q--){
		int op,l,r;
		
		cin >> op >> l >> r;
		//op = rand()%2+1,l = rand()%n+1,r = rand()%n+1;
		//if(l>r)swap(l,r);
		if(op == 1){
		//	for(int i = l;i<=r;i++)aa[i] = (aa[i]+F[i-l+1])%mod;
			update(l,r+1,1,F1[l],F2[l]);
		}else{
			int res =query(l,r+1,1);
		//	int rr = 0;
		//	for(int i = l;i<=r;i++)rr = (rr+aa[i]*aa[i])%mod;
		//	assert(res == rr);
			cout << res << endl;
		}//for(int i = 1;i<=n;i++)cout << query(i,i+1,1) << " ";cout << endl;
	} 
	return 0;
}/*
我们考虑直接把 fj 表示为　fi 和 fi-1 的线性组合
那么我们每个点上对应的值是 k1fi+k2fi-1,
这个东西可以直接维护
维护的东西有点太多了。 
5 2
1 3 5
2 1 5
*/