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++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <iostream>
#include <iterator>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

#define FOR(i,k,n) for(int(i)=(k);(i)<(n);++(i))
#define rep(i,n) FOR(i,0,n)
#define all(v) begin(v),end(v)
#define debug(x) //std::cerr<<#x<<": "<<x<<"\n"
#define debug2(x,y) //std::cerr<<#x<<": "<<x<<", "<<#y<<": "<<y<<"\n"
#define debug3(x,y,z) //std::cerr<<#x<<": "<<x<<", "<<#y<<": "<<y<<", "<<#z<<": "<<z<<"\n"

using ll=long long;
using vi=std::vector<int>;
using vvi=std::vector<vi>;
using vll=std::vector<ll>;
using vvll=std::vector<vll>;
template<typename T> using vvec=std::vector<std::vector<T>>;
template<typename T>
auto make_v(size_t sz){return std::vector<T>(sz);}
template<typename T,typename... Ts>
auto make_v(size_t sz,Ts...ts){return std::vector<decltype(make_v<T>(ts...))>(sz,make_v<T>(ts...));}
template<typename T>
void fill_v(T&var,const T&x){var=x;}
template<typename V,typename T>
void fill_v(V&v,const T&x){for(auto&& w:v){fill_v(w,x);}}

template<typename T> std::ostream& operator<<(std::ostream&s,const std::vector<T>&v){
  int sz=v.size();s<<"\n";rep(i,sz){s<<v[i];if(i<sz-1){s<<"\t";}}s<<"\n";return s;}
template<typename T> std::ostream& operator<<(std::ostream&s,const std::vector<std::vector<T>>&v){
  for(auto&& w:v){s<<w;}return s;}
template<typename T> std::ostream& operator<<(std::ostream&s,const std::deque<T>&v){
  int sz=v.size();s<<"\n";rep(i,sz){s<<v[i];if(i<sz-1){s<<"\t";}}s<<"\n";return s;}
template<typename T> std::ostream& operator<<(std::ostream&s,const std::deque<std::deque<T>>&v){
  for(auto&& w:v){s<<w;}return s;}
template<typename T> std::ostream& operator<<(std::ostream&s, const std::set<T>&v){
  s<<"\n";for(auto&& elm:v){s<<elm<<"\t";}s<<"\n";return s;}

inline void scan(int&a){scanf("%d",&a);}
inline void scan(ll&a){scanf("%lld",&a);}
inline void scan(char&a){scanf(" %c",&a);}
inline void scan(double&a){scanf("%lf",&a);}
inline void scan(std::string&s){char BUF[3000000];scanf(" %s",BUF);s=std::string(BUF);}
template<typename T>
inline void scan(std::vector<T>&v){for(auto&& sv:v){scan(sv);}}
template<typename First,typename...Args>
inline void scan(First&f,Args&...args){scan(f);scan(args...);}
inline void print(int a){printf("%d\n",a);}
inline void print(ll a){printf("%lld\n",a);}
inline void print(double a){printf("%.12f\n",a);}
inline void print(std::string s){std::cout<<s<<"\n";}

using namespace std;

void brute(int l, int r) {
  FOR (x, l, r+1) {
    FOR (y, x, r+1) {
      if (y % x == (y ^ x)) {
        debug2(x, y);
      }
    }
  }
}

class PermComb {
public:
  int mod;
  int sz;
  vll fact;
  vll frev;

  PermComb(int mod, int sz) : mod(mod), sz(sz), fact(sz+1), frev(sz+1) {
    fact[0] = 1;
    rep (i, sz) {
      fact[i+1] = fact[i] * (i+1) % mod;
    }

    rep (i, sz+1) {
      frev[i] = po(fact[i], mod-2);
    }
  }

  ll po(ll k, ll x) {
    ll ret = 1;
    while (x) {
      if (x & 1) {
        ret = ret * k % mod;
      }
      k = k * k % mod;
      x >>= 1;
    }
    return ret;
  }

  ll perm(int n, int k) {
    assert (sz >= n && sz >= k);
    if (n < 0 || k < 0 || n < k) {
      return 0;
    }
    return fact[n] * frev[n-k] % mod;
  }

  ll comb(int n, int k) {
    assert (sz >= n && sz >= k);
    if (n < 0 || k < 0 || n < k) {
      return 0;
    }
    return fact[n] * frev[n-k] % mod * frev[k] % mod;
  }
};


int mod = 1000000007;
vll po2;
vll po3;

int cnt0(ll z) {
  int ans = 0;
  for (int i = 0; ; ++i) {
    if (z >> i) {
      if (((z >> i) & 1) == 0) {
        ans += 1;
      }
    } else {
      break;
    }
  }
  return ans;
}

int cnt1(ll z) {
  int ans = 0;
  for (int i = 0; ; ++i) {
    if (z >> i) {
      if (((z >> i) & 1) == 1) {
        ans += 1;
      }
    } else {
      break;
    }
  }
  return ans - 1;
}

ll calc(ll l, ll r, bool recurse) {
  int bitl = 0;
  int bitr = 0;
  for (int i = 0; ; ++i) {
    if (!(l >> i)) {
      bitl = i;
      break;
    }
  }
  for (int i = 0; ; ++i) {
    if (!(r >> i)) {
      bitr = i;
      break;
    }
  }
  if (bitl == bitr) {
    ll mask = (1 << (bitl-1)) - 1;
    return calc(l & mask, r & mask, true);
  }
  debug2(bitl, bitr);

  ll ans = 0;
  FOR (i, bitl, bitr - 1) {
    ans = (ans + po3[i]) % mod;
  }
  debug(ans);

  for (int i = bitl - 1; i >= 0; --i) {
    ll left_l = (l >> i);
    if (left_l & 1) {
      continue;
    }
    ans = (ans + po2[cnt0(left_l + 1)] * po3[i]) % mod;
  }
  ans = (ans + po2[cnt0(l)]) % mod;
  debug(ans);

  for (int i = bitr - 2; i >= 0; --i) {
    ll left_r = (r >> i);
    if ((left_r & 1) == 0) {
      continue;
    }
    ans = (ans + po2[cnt1(left_r - 1)] * po3[i]) % mod;
  }
  if (recurse && bitl == 0) {
  debug(ans);
    ans = (ans + po2[cnt1(r) + 1]) % mod;
  } else {
    debug(po2[cnt1(r)]);
    ans = (ans + po2[cnt1(r)]) % mod;
  }
  debug(ans);

  return ans;
}

int main() {
  // \sum_{k=1}^{i}
  int m = 101;
  po2.assign(m, 1);
  po3.assign(m, 1);
  FOR (i, 0, m-1) {
    po2[i+1] = po2[i] * 2 % mod;
    po3[i+1] = po3[i] * 3 % mod;
  }

  ll l, r;
  scan(l, r);

  ll ans = calc(l, r, false);
  print(ans);

  return 0;
}