from functools import lru_cache

def General_Binary_Decrease_Search_Integer(L, R, cond, default=None):
    """ 条件式が単調減少であるとき, 整数上で二部探索を行う.

    L: 解の下限
    R: 解の上限
    cond: 条件 (1変数関数, 広義単調減少 を満たす)
    default: R で条件を満たさないときの返り値
    """

    if not(cond(L)):
        return default

    if cond(R):
        return R

    L-=1
    while R-L>1:
        C=L+(R-L)//2
        if cond(C):
            L=C
        else:
            R=C
    return L

@lru_cache(maxsize=None)
def count(N):
    if N<10:
        Ans=0
        for n in range(1,N+1):
            Ans+=A[n]
        return Ans

    q,r=divmod(N,10)
    x=y=0
    for i in range(10):
        if i<=r:
            x+=A[i]
        else:
            y+=A[i]
    return count(q)*(r+1)+x*(q+1)-1+count(q-1)*(9-r)+y*q
#==================================================

A=[1,0,0,0,1,0,1,0,2,1]
K=int(input())

N=General_Binary_Decrease_Search_Integer(0,10**18,lambda x:count(x)<=K)
print(N if count(N)==K else -1)