# 最下位ビットが0
def f(n: int, k: int) -> int:
    b = 2 ** (k+1)
    cnt = (n + 1) // b

    a = 2**k * (2**k + (2**(k+1) - 1)) // 2
    d = 2 ** (2*(k+1) - 1)
    s = (a + a+d*(cnt-1)) * cnt // 2

    p1 = cnt * b + b//2
    p2 = min(n, p1 + (b // 2))
    # print(f'{p1, p2}')
    if p1 <= p2:
        return s + (p1 + p2) * (p2 - p1 + 1) // 2

    return s


MOD = 10**9 + 7
N = int(input())
ans = 0
for i in range(60):
    ans += f(N, i)
    ans %= MOD

print(ans)