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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#define size 100
#define p (int)(1e9 + 7)

int main(){
	int i;
	long long int K[size], L1[size], L2[size], L3[size];
	long long int n;
	scanf("%lld", &n);

	for(i = 0; n != 0; i++){
		K[i] = n;
		n = (n-1)/2;
	}
	K[i]  = 1;
	L1[i] = 1;
	L2[i] = 1;
	L3[i] = 0;
	i--;

	for(; i >= 0; i--){
		if(K[i] % 2 == 0){
			K[i] = (3*K[i + 1] + L1[i + 1])%p;
			L1[i] = L1[i + 1];
			L2[i] = (L1[i + 1] + L2[i + 1])%p;
			L3[i] = L2[i + 1];
		}
		else{
			K[i] = (3*K[i + 1] - L2[i + 1] + p)%p;
			L1[i] = (L1[i + 1] + L2[i + 1])%p;
			L2[i] = L2[i + 1];
			L3[i] = (L2[i + 1] + L3[i + 1])%p;
		}
	}
	printf("%lld\n", K[0]);
	return 0;
}