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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#define SF scanf
#define PF printf
#define MAXN 1000010
#define INF 0x3FFFFFFF
#define MOD 1000000007
using namespace std;
typedef long long ll;
ll n,n1;
ll a[MAXN];
map<ll,ll> mp;
ll p(ll x){
	if(mp.count(x))
		return mp[x];
	if(x<=1000000)
		return a[x];
	if(x==1)
		return 1;
	if(x%2ll==1ll)
		mp[x]=(p(x/2ll)+p(x/2ll+1ll))%MOD;
	else
		mp[x]=p(x/2ll);
	return mp[x];
}
ll que(ll x){
	if(x==1)
		return 1;
	if(x%2ll==1ll)
		return (que(x/2ll)*3ll+p(x/2ll+1ll))%MOD;
	else
		return (que(x/2ll)*3ll-p(x/2ll)+MOD)%MOD;
}
void prepare(){
	a[1]=1;
	for(int i=1;i<=1000000;i++){
		if(i%2==1)
			a[i]=(a[i/2]+a[i/2+1])%MOD;	
		else
			a[i]=a[i/2];
	}
}	
int main(){
	prepare();
	SF("%lld",&n1);
	n1++;
	PF("%lld\n",que(n1));
}