# 前処理.
MAX_BIT = 30
U = [2*(pow(4,k)-1)//3 for k in range(MAX_BIT + 1)]

def h(n, v):
    if n == 0: return 0
    m = n - 1
    memo = {}
    def dfs(d, fx, fy):
        if d == -1: return 1
        st = (d, fx, fy)
        if st in memo: return memo[st]
        res = 0
        bm = (m >> d) & 1
        bv = (v >> d) & 1
        for bx in (0, 1):
            for by in (0, 1):
                if fx == 0 and bx > bm: continue
                if fy == 0 and by > bm: continue
                ok = 0
                if d & 1:
                    if (bx | by) == bv:
                        ok = 1
                else:
                    if (bx & by) == bv:
                        ok = 1
                if ok:
                    res += dfs(d-1, fx or (bx < bm), fy or (by < bm))
        memo[st] = res
        return res
    return dfs(30, 0, 0)

N = int(input())
ans = 0
left = 1
for k in range(1, MAX_BIT + 1):
    l = left
    r = N+1
    # g(n) = U[k]となる境界nを探す.
    while r-l > 1:
        mid = (l+r)//2
        if h(mid, U[k - 1]) >= h(mid, U[k]):
            l = mid
        else:
            r = mid
    ans += (r-left)*U[k-1]
    left = r
print(ans)