The final match of the Berland Football Cup has been held recently. The referee has shown n yellow cards throughout the match. At the beginning of the match there were a_1 players in the first team and a_2 players in the second team.

The rules of sending players off the game are a bit different in Berland football. If a player from the first team receives k_1 yellow cards throughout the match, he can no longer participate in the match — he's sent off. And if a player from the second team receives k_2 yellow cards, he's sent off. After a player leaves the match, he can no longer receive any yellow cards. Each of n yellow cards was shown to exactly one player. Even if all players from one team (or even from both teams) leave the match, the game still continues.

The referee has lost his records on who has received each yellow card. Help him to determine the minimum and the maximum number of players that could have been thrown out of the game.

Input

The first line contains one integer a_1 (1 ≤ a_1 ≤ 1 000) — the number of players in the first team.

The second line contains one integer a_2 (1 ≤ a_2 ≤ 1 000) — the number of players in the second team.

The third line contains one integer k_1 (1 ≤ k_1 ≤ 1 000) — the maximum number of yellow cards a player from the first team can receive (after receiving that many yellow cards, he leaves the game).

The fourth line contains one integer k_2 (1 ≤ k_2 ≤ 1 000) — the maximum number of yellow cards a player from the second team can receive (after receiving that many yellow cards, he leaves the game).

The fifth line contains one integer n (1 ≤ n ≤ a_1 ⋅ k_1 + a_2 ⋅ k_2) — the number of yellow cards that have been shown during the match.

Output

Print two integers — the minimum and the maximum number of players that could have been thrown out of the game.

Examples

Input


2
3
5
1
8


Output


0 4


Input


3
1
6
7
25


Output


4 4


Input


6
4
9
10
89


Output


5 9

Note

In the first example it could be possible that no player left the game, so the first number in the output is 0. The maximum possible number of players that could have been forced to leave the game is 4 — one player from the first team, and three players from the second.

In the second example the maximum possible number of yellow cards has been shown (3 ⋅ 6 + 1 ⋅ 7 = 25), so in any case all players were sent off.
''' CODED WITH LOVE BY SATYAM KUMAR '''

from sys import stdin, stdout
import heapq
import cProfile, math
from collections import Counter, defaultdict, deque
from bisect import bisect_left, bisect, bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
import operator as op
from functools import reduce
import sys

sys.setrecursionlimit(10 ** 6)  # max depth of recursion
threading.stack_size(2 ** 27)  # new thread will get stack of such size
fac_warm_up = False
printHeap = str()
memory_constrained = False
P = 10 ** 9 + 7


class MergeFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n
        self.num_sets = n
        self.lista = [[_] for _ in range(n)]

    def find(self, a):
        to_update = []
        while a != self.parent[a]:
            to_update.append(a)
            a = self.parent[a]
        for b in to_update:
            self.parent[b] = a
        return self.parent[a]

    def merge(self, a, b):
        a = self.find(a)
        b = self.find(b)
        if a == b:
            return
        if self.size[a] < self.size[b]:
            a, b = b, a
        self.num_sets -= 1
        self.parent[b] = a
        self.size[a] += self.size[b]
        self.lista[a] += self.lista[b]

    def set_size(self, a):
        return self.size[self.find(a)]

    def __len__(self):
        return self.num_sets


def display(string_to_print):
    stdout.write(str(string_to_print) + "\n")


def prime_factors(n):  # n**0.5 complex
    factors = dict()
    for i in range(2, math.ceil(math.sqrt(n)) + 1):
        while n % i == 0:
            if i in factors:
                factors[i] += 1
            else:
                factors[i] = 1
            n = n // i
    if n > 2:
        factors[n] = 1
    return (factors)


def all_factors(n):
    return set(reduce(list.__add__,
                      ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))


def fibonacci_modP(n, MOD):
    if n < 2: return 1
    return (cached_fn(fibonacci_modP, (n + 1) // 2, MOD) * cached_fn(fibonacci_modP, n // 2, MOD) + cached_fn(
        fibonacci_modP, (n - 1) // 2, MOD) * cached_fn(fibonacci_modP, (n - 2) // 2, MOD)) % MOD


def factorial_modP_Wilson(n, p):
    if (p <= n):
        return 0
    res = (p - 1)
    for i in range(n + 1, p):
        res = (res * cached_fn(InverseEuler, i, p)) % p
    return res


def binary(n, digits=20):
    b = bin(n)[2:]
    b = '0' * (digits - len(b)) + b
    return b


def is_prime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True


def generate_primes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    return prime


factorial_modP = []


def warm_up_fac(MOD):
    global factorial_modP, fac_warm_up
    if fac_warm_up: return
    factorial_modP = [1 for _ in range(fac_warm_up_size + 1)]
    for i in range(2, fac_warm_up_size):
        factorial_modP[i] = (factorial_modP[i - 1] * i) % MOD
    fac_warm_up = True


def InverseEuler(n, MOD):
    return pow(n, MOD - 2, MOD)


def nCr(n, r, MOD):
    global fac_warm_up, factorial_modP
    if not fac_warm_up:
        warm_up_fac(MOD)
        fac_warm_up = True
    return (factorial_modP[n] * (
                (pow(factorial_modP[r], MOD - 2, MOD) * pow(factorial_modP[n - r], MOD - 2, MOD)) % MOD)) % MOD


def test_print(*args):
    if testingMode:
        print(args)


def display_list(list1, sep=" "):
    stdout.write(sep.join(map(str, list1)) + "\n")


def display_2D_list(li):
    for i in li:
        print(i)


def prefix_sum(li):
    sm = 0
    res = []
    for i in li:
        sm += i
        res.append(sm)
    return res


def get_int():
    return int(stdin.readline().strip())


def get_tuple():
    return map(int, stdin.readline().split())


def get_list():
    return list(map(int, stdin.readline().split()))


memory = dict()


def clear_cache():
    global memory
    memory = dict()


def cached_fn(fn, *args):
    global memory
    if args in memory:
        return memory[args]
    else:
        result = fn(*args)
        memory[args] = result
        return result


def ncr(n, r):
    return math.factorial(n) / (math.factorial(n - r) * math.factorial(r))


def binary_search(i, li):
    fn = lambda x: li[x] - x // i
    x = -1
    b = len(li)
    while b >= 1:
        while b + x < len(li) and fn(b + x) > 0:  # Change this condition 2 to whatever you like
            x += b
        b = b // 2
    return x


# -------------------------------------------------------------- MAIN PROGRAM


TestCases = False
fac_warm_up_size = 10 ** 5 + 100
optimise_for_recursion = False  # Can not be used clubbed with TestCases WHen using recursive functions, use Python 3


def main():
    a1 = get_int()
    a2 = get_int()
    k1 = get_int()
    k2 = get_int()
    n = get_int()
    if k2>k1:
        a1,a2,k1,k2 = a2,a1,k2,k1
    mini = max(0, n - a1*(k1-1) - a2*(k2-1))
    if n<= a2*k2:
        maxi = n//k2
    else:
        maxi = a2 + (n-a2*k2)//k1
    print(mini, maxi)
# --------------------------------------------------------------------- END=


if TestCases:
    for i in range(get_int()):
        main()
else:
    main() if not optimise_for_recursion else threading.Thread(target=main).start()