Problem:
Score : 600 points

Problem Statement

You are given a positive integer N.
Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v.
Here, xor denotes the bitwise exclusive OR.
Since it can be extremely large, compute the answer modulo 10^9+7.

Constraints

1≦N≦10^{18}

Input

The input is given from Standard Input in the following format:

N

Output

Print the number of the possible pairs of integers u and v, modulo 10^9+7.

Sample Input 1

3

Sample Output 1

5

The five possible pairs of u and v are:

u=0,v=0 (Let a=0,b=0, then 0 xor 0=0, 0+0=0.)

u=0,v=2 (Let a=1,b=1, then 1 xor 1=0, 1+1=2.）

u=1,v=1 (Let a=1,b=0, then 1 xor 0=1, 1+0=1.）

u=2,v=2 (Let a=2,b=0, then 2 xor 0=2, 2+0=2.）

u=3,v=3 (Let a=3,b=0, then 3 xor 0=3, 3+0=3.）

Sample Input 2

1422

Sample Output 2

52277

Sample Input 3

1000000000000000000

Sample Output 3

787014179

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int findPair(int n)
{
	int a,b,counter=0,u,v,temp1,temp2,temp3,temp4;

	for(u=0;u<=n;u++)
	{
		for(v=0;v<=n;v++)
		{
			for(a=0;a<=n;a++)
			{
				for(b=0;b<=n;b++)
				{
					temp1=a^b;
					temp2=a+b;
					if(temp3==temp1 && temp4==temp2)
					{
						break;
					}
					if(temp1==u && temp2==v)
					{
						temp3=temp1;
						temp4=temp2;
						counter++;
					}
				}
			}
		}
	}
	return counter;
}

int main(int argc, char *argv[]) {
	int n;

	scanf("%d",&n);
	printf("%d",findPair(n));
	return 0;
}