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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

public class Main {
	Scanner sc;
	long L, R;
	
	private static long MOD = 1000000007L;
	private static final long[] MASK;
	static {
		MASK = new long[64];
		MASK[0] = 0L;
		for (int i = 1; i < 64; i++) MASK[i] = MASK[i-1] * 2 + 1;
	}
	private static final long[] POW3;
	static {
		POW3 = new long[64];
		POW3[0] = 1L;
		for (int i = 1; i < 64; i++) POW3[i] = (POW3[i-1] * 3L) % MOD;
	}
	
	public Main() {
		sc = new Scanner(System.in);
	}
	
	private void calc() {
		L = sc.nextLong();
		R = sc.nextLong();
		
		System.out.println(count(0, 0, 63));
		System.out.flush();
	}
	
	private long count(long x, long y, int maskn) {
		if ( (x|y) != 0 && y/x > 1) return 0L; //(y>>>1) >= x) return 0L;
		if (R < x || R < y) return 0L;
		if ((y | MASK[maskn]) < L || (x | MASK[maskn]) < L) return 0L;
		if (L <= x && (y | MASK[maskn]) <= R) return POW3[maskn];
		maskn--;
		long nextBit = MASK[maskn]+1L;
		long su = count(x | nextBit, y | nextBit, maskn);
		long sl = count(x, y, maskn);
		if (x == y) return (su + sl + count(x, y | nextBit, maskn)) % MOD;
		if ( (L & (~MASK[maskn])) < x ) return (2 * su + sl) % MOD;
		if ( y < (R | MASK[maskn]) ) return (su + 2 * sl) % MOD;
		return (su + sl + count(x, y | nextBit, maskn)) % MOD;
	}
	
	public static void main(String[] args) {
		new Main().calc();
	}
}