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<bits/stdc++.h>
using namespace std;
#define sync ios_base::sync_with_stdio(false), cin.tie(NULL)
#define F first
#define S second
#define pb emplace_back
#define mt make_tuple
#define mp make_pair
#define gcd __gcd
#define PI 3.141592653589
// Input
#define in(a) scanf("%d",&a)
#define in2(a,b) scanf("%d%d",&a,&b)
#define in3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define llin(a) cin >> a
#define inl(a) scanf("%lld",&a)
#define read(v,i,n) for(i=0;i<n;i++)in(v[i])
#define twod(mat,i,j,n,m) rep(i,n){rep(j,m)in(mat[i][j]);}
#define sc(ch) scanf("%c",&ch)
#define sstr(str) scanf("%s",str)
// Output
#define pr(a) printf("%d ",a)
#define pr2(a,b) printf("%d %d\n",a,b)
#define pr3(a,b,c) printf("%d %d %d\n",a,b,c)
#define out(a) printf("%d\n",a)
#define outl(a) printf("%lld\n",a)
#define llpr(a) cout << a << " "
#define llout(a) cout << a << "\n"
#define yes printf("YES\n")
#define no printf("NO\n")
#define lin printf("\n")
// Iterator
#define lp(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define all(vec) vec.begin(),vec.end()
#define lower(v,k) lower_bound(v.begin(),v.end(),k)-v.begin()
#define upper(v,k) upper_bound(v.begin(),v.end(),k)-v.begin()
#define tf(mytuple) get<0>(mytuple)
#define ts(mytuple) get<1>(mytuple)
#define tt(mytuple) get<2>(mytuple)
// Debug
#define dbg(v,i,n) for(i=0;i<n;i++)pr(v[i]); lin
#define what(x) cerr << #x << " : " << x << "\n"
#define ck printf("continue\n")
#define debug(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args)
{
	cerr << *it << " = " << a << "\n";
	err(++it, args...);
}
// Data Type
#define ll long long int
#define ii pair<int,int>
#define pli pair<ll,int>
#define triple tuple<int,int,int>
#define vi vector<int>
#define vii vector<pair<int,int> >
#define vvi vector<vector<int> >
#define viii vector<pair<pair<int,int>,int> >
#define vvii vector<vector<pair<int,int> > >
#define M 1000000007
#define N 61  //check the limits, man
const ll INF = 1e18;
ll dp[N][2][2];
int bit[N];
ll fun(int i,int mod,int ok)
{
    if(i==N)
        return ((!mod)&ok);
    if(dp[i][mod][ok]>=0)
        return dp[i][mod][ok];
    dp[i][mod][ok]=0;
    int a,b;
    rep(a,2)
    {
        rep(b,a+1)
        {
            int cur=((a+b+mod)&1);
            int nmod=(a+b+mod)/2;
            int nok=ok;
            if(bit[i]<cur)nok=0;
            if(bit[i]>cur)nok=1;
            dp[i][mod][ok]+=fun(i+1,nmod,nok);
        }
    }
    dp[i][mod][ok]%=M;
    return dp[i][mod][ok];
}
int main()
{
    ll n;
    cin >> n;
    int i;
    rep(i,N)
        bit[i]=(n&1),n/=2;
    rep(i,N)
        dp[i][0][0]=-1,dp[i][0][1]=-1,dp[i][1][0]=-1,dp[i][1][1]=-1;
    llout(fun(0,0,1));
}