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;
ll n, mod = 1e9+7, dp[61][3];

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n;
	dp[0][0] = 1;
	for(int i=1; i<=60; ++i){
		int t = 60 - i;
		if(((n >> t) & 1) == 0){
			dp[i][0] = dp[i-1][0] + dp[i-1][1];
			dp[i][1] = dp[i-1][1];
			dp[i][2] = dp[i-1][1] + 3 * dp[i-1][2];
			for(int j=0; j<3; ++j)
				dp[i][j] %= mod;
		}else{
			dp[i][0] = dp[i-1][0];
			dp[i][1] = dp[i-1][0] + dp[i-1][1];
			dp[i][2] = 2 * dp[i-1][1] + 3 * dp[i-1][2];
			for(int j=0; j<3; ++j)
				dp[i][j] %= mod;
		}
	}
//	for(int i=0; i<61; ++i)
//		cout << dp[i][0] << " " << dp[i][1] << " " << dp[i][2] << "\n";
	cout << (dp[60][0] + dp[60][1] + dp[60][2]) % mod << "\n";
}