Fox Ciel studies number theory.

She thinks a non-empty set S contains non-negative integers is perfect if and only if for any $a, b \in S$ (a can be equal to b), $(a \text{xor} b) \in S$. Where operation xor means exclusive or operation (http://en.wikipedia.org/wiki/Exclusive_or).

Please calculate the number of perfect sets consisting of integers not greater than k. The answer can be very large, so print it modulo 1000000007 (10^9 + 7).


-----Input-----

The first line contains an integer k (0 ≤ k ≤ 10^9).


-----Output-----

Print a single integer — the number of required sets modulo 1000000007 (10^9 + 7).


-----Examples-----
Input
1

Output
2

Input
2

Output
3

Input
3

Output
5

Input
4

Output
6



-----Note-----

In example 1, there are 2 such sets: {0} and {0, 1}. Note that {1} is not a perfect set since 1 xor 1 = 0 and {1} doesn't contain zero.

In example 4, there are 6 such sets: {0}, {0, 1}, {0, 2}, {0, 3}, {0, 4} and {0, 1, 2, 3}.

Difficulty: competition

Starter code:

Solutions:
from math import factorial
MOD = 10**9+7

k = int(input())
bink = list(map(int, bin(k)[2:]))
N = len(bink)

# dp[i][j][k] = first i bits, j bases,
#   k = 1 if maxor matches k, 0 else
dp = [[[0,0] for j in range(i+2)] for i in range(N+1)]
dp[0][0][1] = 1
for i in range(1, N+1):
    for j in range(i+1):
        # k = 0 -> 0
        dp[i][j][0]+= 2**j * dp[i-1][j][0] # distribute
        if j: dp[i][j][0]+= dp[i-1][j-1][0] # new base
        # k = 1 -> 0
        odd = 2**(j-1) if j else 0
        even = 2**j - odd
        if bink[i-1] == 1:
            dp[i][j][0]+= even * dp[i-1][j][1] # distribute even
        # k = 1 -> 1
        if bink[i-1] == 0:
            dp[i][j][1]+= even * dp[i-1][j][1] # distribute even
        else:
            dp[i][j][1]+= odd * dp[i-1][j][1] # distribute odd
            if j: dp[i][j][1]+= dp[i-1][j-1][1] # new base
ans = sum(map(sum, dp[-1]))
print(ans % MOD)