Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3

Sample Output 1

5

The five possible pairs of u and v are:

u=0,v=0 (Let a=0,b=0, then 0 xor 0=0, 0+0=0.)

u=0,v=2 (Let a=1,b=1, then 1 xor 1=0, 1+1=2.）

u=1,v=1 (Let a=1,b=0, then 1 xor 0=1, 1+0=1.）

u=2,v=2 (Let a=2,b=0, then 2 xor 0=2, 2+0=2.）

u=3,v=3 (Let a=3,b=0, then 3 xor 0=3, 3+0=3.）

Sample Input 2

1422

Sample Output 2

52277

Sample Input 3

1000000000000000000

Sample Output 3

787014179

Solution in Ruby (2.3.3) (judge verdict: Accepted)

N = gets.to_i
NS = N.to_s(2)
NL = NS.length
MOD = 10**9 + 7

dp = Array.new(NL){Array.new(3, 0)}

dp[0][0] = 1
dp[0][1] = 1

(1...NL).each do |i|
    if NS[i] == '1'
        dp[i][0] = dp[i - 1][0]
        dp[i][1] = dp[i - 1][0] + dp[i - 1][1]
        dp[i][2] = dp[i - 1][1] * 2 + dp[i - 1][2] * 3
    else
        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] + dp[i - 1][2] * 3
    end
    dp[i][0] %= MOD
    dp[i][1] %= MOD
    dp[i][2] %= MOD
end

puts dp[NL - 1].inject(:+) % MOD