Problem:
Score : 600 points

Problem Statement

Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

1 \leq L \leq R \leq 10^{18}

Input

Input is given from Standard Input in the following format:

L R

Output

Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.

Sample Input 1

2 3

Sample Output 1

3

Three pairs satisfy the condition: (2, 2), (2, 3), and (3, 3).

Sample Input 2

10 100

Sample Output 2

604

Sample Input 3

1 1000000000000000000

Sample Output 3

68038601

Be sure to compute the number modulo 10^9 + 7.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define all(a)a.begin(),a.end()
using ll=long long;
const int INF = 1<<30;
const ll INFll =1LL<<62;
const int mod= int(1e9)+7;
//const int mod=998244353;
using P = pair<int,int>;
using Pl= pair<ll,ll>;
using ld=long double;
using V=vector<int>;
using Vl=vector<ll>;
using VV=vector<vector<int>>;
using VVl=vector<vector<ll>>;

// modint: mod 計算を int を扱うように扱える構造体
template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};
using mint = Fp<mod>;

const int n=60;
mint dp[n+1][2][2][2]={};

int main(){
    dp[0][0][0][0]=1;
    ll L,R;cin >>L>>R;
    for (int i =0; i<n; ++i) {
        int Li=(L>>(n-1-i))&1;
        int Ri=(R>>(n-1-i))&1;
        for (int j = 0; j < 2; ++j) {
            for (int l= 0; l < 2; ++l) {
                for (int r = 0; r < 2; ++r) {
                    for (int x = 0; x < 2; ++x) {
                        for (int y = 0; y < 2; ++y) {
                            int j2=j,l2=l,r2=r;
                            if(x>y)continue;
                            if(j==0 and y==1 and x==0)continue;
                            if(j==0 and x==1 and y==1)j2=1;
                            if(l==0 and x<Li)continue;
                            if(l==0 and x>Li)l2=1;
                            if(r==0 and y>Ri)continue;
                            if(r==0 and y<Ri)r2=1;
                            dp[i+1][j2][l2][r2]+=dp[i][j][l][r];
                        }
                    }
                }
            }
        }
    }
    mint ans=0;
    for (int j = 0; j < 2; ++j) {
        for (int l = 0; l < 2; ++l) {
            for (int r = 0; r < 2; ++r) {
                ans+=dp[n][j][l][r];
            }
        }
    }
    cout <<ans <<endl;
}