https://codeforces.com/contest/1151/problem/C 

题解：前缀和+等差数列

1、认识到这是一个等差数列；

2、预处理出每一段的第一项、每一段和、每一段前缀和；

3、减少分类情况采用左区间固定1的方法，即[l，r]=[1，r]-[1，l-1]；

4、用log2函数判断当前端点第几段；

5、特判右端点0和第0段；

6、完整的段数前缀和+右端点所在的非完整的和；

7、注意求模；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp;
ll sum[N];
ll f[N],pow2[N];
char str;
struct node{};
void init(){
    bool flag=1;
    ll base=1;
    ll odd=1;
    ll even=2;
    sum[0]=1;
    for(int i=0;i<=63;i++){
        pow2[i]=base;
        if(flag){
            f[i]=odd;
            odd+=2*base;
        }else{
            f[i]=even;
            even+=2*base;
        }
        if(i)sum[i]=(sum[i-1]+((f[i]%MOD)*(base%MOD))%MOD+((base%MOD)*((base-1)%MOD))%MOD)%MOD;
        flag=!flag;
        base*=2;
    }
}
ll sloved(ll R){
    if(R<=0)
        return 0;
    int r=log2(R);//最后一段指数
    if(r==0)
        return 1;
    ll cnt=R-pow2[r]+1;//cout<<sum[r]<<endl;
    return (sum[r-1]+((f[r]%MOD)*(cnt%MOD))%MOD+((cnt%MOD)*((cnt-1)%MOD))%MOD)%MOD;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    init();
    scanf("%I64d%I64d",&l,&r);
    cout<<(sloved(r)-sloved(l-1)+MOD)%MOD<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}