#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>

using namespace std;

#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define LINF (ll)INF*INF
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<(n);i++)
#define loop(i,a,n) for(int i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

#define int ll //!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;
typedef vector<pii> vp;

int gcd(int a, int b){
    if(b==0) return a;
    return gcd(b,a%b);
}
int lcm(int a, int b){
    return a/gcd(a,b)*b;
}

int dp[60][60][2][2];
//i桁目までみて、popcountがjで、N以下が確定している(k=1)のとき、
//dp[i][j][k][0] : そのような数の通り数
//dp[i][j][k][1] : そのような数の総和

signed main(void) {
    int n;
    cin >> n;
    string s = "";
    while(n){
        s = char('0'+n%2) + s;
        n /= 2;
    }
    n = s.size();
    dp[0][0][0][0] = 1;
    rep(i,n)rep(j,n)rep(k,2){
        int lim = 1;
        if(k == 0 && s[i] == '0')lim = 0;
        rep(x,lim+1){
            (dp[i+1][j+x][k || x<lim][0] += dp[i][j][k][0]) %= MOD;
            (dp[i+1][j+x][k || x<lim][1] += (dp[i][j][k][1]*2 + dp[i][j][k][0]*x)%MOD) %= MOD;
        }
    }
    int ans = 0;
    rep(j,n)rep(k,2)ans = (ans + dp[n][j][k][1]*j) % MOD;
    cout << ans << endl;
}