#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
const ll inf = 1e9;

const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

mint dp[65][3][2];

int main() {
  const int dig=60;
  ll n;
  cin>>n;
  dp[0][0][0]=mint(1);
  rep(i,dig) rep(j,3) rep(k,2){
    mint pre=dp[i][j][k];
    if(pre.x==0) continue;
    int a=(n>>(dig-1-i))&1;
    // cout<<"i="<<i<<" j="<<j<<" k="<<k<<" pre="<<pre<<" a="<<a<<endl;
    if(j==0 && k==0 && a==0) dp[i+1][0][0]+=pre;
    if(j==0 && k==0 && a==1){ 
      dp[i+1][0][1]+=pre;
      dp[i+1][1][0]+=pre;
    }
    if(j==0 && k==1){ 
      dp[i+1][0][1]+=pre;
      dp[i+1][1][1]+=pre;
    }
    if(j>0 && k==0 && a==0) dp[i+1][j][0]+=pre*2;
    if(j==1 && k==0 && a==1){ 
      dp[i+1][1][0]+=pre;
      dp[i+1][1][1]+=pre*2;
      dp[i+1][2][0]+=pre;
    }
    if(j==1 && k==1){ 
      dp[i+1][1][1]+=pre*3;
      dp[i+1][2][1]+=pre;
    }
    if(j==2 && k==0 && a==1){ 
      dp[i+1][2][0]+=pre*2;
      dp[i+1][2][1]+=pre*2;
    }
    if(j==2 && k==1) dp[i+1][2][1]+=pre*4;
    // rep(j,3) rep(k,2) cout<<dp[i+1][j][k]<<" ";
    // cout<<endl;
  } 
  mint ans=dp[dig][2][0]+dp[dig][2][1];
  cout<<ans<<endl;
  return 0;
}