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.

Examples

Input

3


Output

5


Input

1422


Output

52277


Input

1000000000000000000


Output

787014179
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<ll,ll> P;
#define pb push_back
#define ft first
#define sd second
#define mp make_pair
#define fr(i,n) for(int i=0;i<n;i++)
#define Fr(i,n) for(int i=0;i++<n;)
#define ifr(i,n) for(int i=n-1;i>=0;i--)
#define iFr(i,n) for(int i=n;i>0;i--)

const ll MOD=1e9+7;

ll ad(ll x,ll y=0){
  return (x+y)%MOD;
}

int main(){
  ll n,dp[70][3]={};
  cin>>n;
  dp[62][0]=1;
  ifr(i,62){
    if((n>>i)&1){
      dp[i][0]=dp[i+1][0];
      dp[i][1]=ad(dp[i+1][0],dp[i+1][1]);
      dp[i][2]=ad(2*dp[i+1][1],3*dp[i+1][2]);
    }
    else{
      dp[i][0]=ad(dp[i+1][0],dp[i+1][1]);
      dp[i][1]=dp[i+1][1];
      dp[i][2]=ad(dp[i+1][1],3*dp[i+1][2]);
    }
  }
  cout<<ad(dp[0][0]+dp[0][1]+dp[0][2])<<endl;
}