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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
	M,L,R,ans,ll,rr:int64;
    k,i,x,y,xl,yl,limx,limy,xx,yy,id,ie:Longint;
    dp:array[1..64,1..2,1..2]of int64;
begin
	read(L,R);
    M:=1000000007;
    ll:=1;
    rr:=2;
    for k:=1 to 62 do begin
    	if(rr<=L)or(R<ll)then begin
        	ll:=ll*2;
            rr:=rr*2;
        	continue;
        end;
        for i:=1 to 64 do for xx:=1 to 2 do for yy:=1 to 2 do dp[i,xx,yy]:=0;
        if L<=ll then if rr-1<=R then dp[1,2,2]:=1 else dp[1,2,1]:=1 else if rr-1<=R then dp[1,1,2]:=1 else dp[1,1,1]:=1;
        for i:=1 to k-1 do begin
        	x:=(L>>(k-1-i))mod 2;
            y:=(R>>(k-1-i))mod 2;
            for xl:=1 to 2 do begin
            	if xl=1 then limx:=x else limx:=0;
                for yl:=1 to 2 do begin
                	if yl=1 then limy:=y else limy:=1;
                    for yy:=0 to limy do begin
                    	for xx:=limx to yy do begin
                        	if(xl=2)or(xx>limx)then id:=2 else id:=1;
                            if(yl=2)or(yy<limy)then ie:=2 else ie:=1;
                            dp[i+1,id,ie]:=(dp[i+1,id,ie]+dp[i,xl,yl])mod M;
                        end;
                    end;
                end;
            end;
        end;
        ll:=ll*2;
        rr:=rr*2;
        ans:=(ans+dp[k,1,1]+dp[k,1,2]+dp[k,2,1]+dp[k,2,2])mod M;
    end;
    writeln(ans);
end.