Buggy python submission (verdict: Time Limit Exceeded):
import math
import os
import sys
from functools import lru_cache

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7

N, S = [int(sys.stdin.readline()) for _ in range(2)]


@lru_cache(maxsize=None)
def f(b, n):
    if n < b:
        return n
    return f(b, int(n // b)) + n % b


def bisect_left_callable(fn, x, lo, hi):
    """
    lo から hi-1 のうち、fn の結果が x 以上となる、最も左の値
    bisect.bisect_left と同じ
    https://docs.python.org/ja/3/library/bisect.html
    :param callable fn:
    :param x:
    :param int lo: 最小値
    :param int hi: 最大値 + 1
    :return: lo <= ret <= hi
    """
    while lo < hi:
        mid = (lo + hi) // 2
        if fn(mid) < x:
            lo = mid + 1
        else:
            hi = mid
    return lo


if S > N:
    print(-1)
    exit()
if S == N:
    print(N + 1)
    exit()

# #########
# b = 2
# ans = -1
# prev = 0
# while b <= N:
#     if f(b, N) > prev:
#         print(b, N, f(b, N))
#     prev = f(b, N)
#     b += 1
# ##########

b = 2
ans = -1
while b <= math.sqrt(N) + 1:
    if f(b, N) == S:
        ans = b
        break
    b += 1
if ans >= 0:
    print(ans)
    exit()

for i in range(1, int(math.sqrt(N)) + 1):
    hi = N // i + 1
    lo = N // (i + 1) + 1
    b = bisect_left_callable(lambda x: -f(x, N), -S, lo, hi)
    if f(b, N) == S:
        ans = b
print(ans)



Fixed submission:
import math
import os
import sys
from functools import lru_cache

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7

N, S = [int(sys.stdin.readline()) for _ in range(2)]


@lru_cache(maxsize=None)
def f(b, n):
    if n < b:
        return n
    return f(b, int(n // b)) + n % b


if S > N:
    print(-1)
    exit()
if S == N:
    print(N + 1)
    exit()

# #########
# b = 2
# ans = -1
# prev = 0
# while b <= N:
#     if f(b, N) > prev:
#         print(b, N, f(b, N))
#     prev = f(b, N)
#     b += 1
# ##########

b = 2
ans = -1
while b <= math.sqrt(N) + 1:
    if f(b, N) == S:
        ans = b
        break
    b += 1
if ans >= 0:
    print(ans)
    exit()

ans = INF
# b > sqrt(N) のとき 2 桁になるので上位桁を全探索する
for d2 in range(1, int(math.sqrt(N)) + 1):
    if (N - S) % d2 == 0:
        b = int(1 + (N - S) // d2)
        if b >= 2 and f(b, N) == S:
            ans = min(ans, b)
if ans == INF:
    print(-1)
else:
    print(ans)

Buggy python submission (verdict: Time Limit Exceeded):
import math
import os
import sys
from functools import lru_cache

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7

N, S = [int(sys.stdin.readline()) for _ in range(2)]


@lru_cache(maxsize=None)
def f(b, n):
    if n < b:
        return n
    return f(b, int(n // b)) + n % b


def bisect_left_callable(fn, x, lo, hi):
    """
    lo から hi-1 のうち、fn の結果が x 以上となる、最も左の値
    bisect.bisect_left と同じ
    https://docs.python.org/ja/3/library/bisect.html
    :param callable fn:
    :param x:
    :param int lo: 最小値
    :param int hi: 最大値 + 1
    :return: lo <= ret <= hi
    """
    while lo < hi:
        mid = (lo + hi) // 2
        if fn(mid) < x:
            lo = mid + 1
        else:
            hi = mid
    return lo


if S > N:
    print(-1)
    exit()
if S == N:
    print(N + 1)
    exit()

# #########
# b = 2
# ans = -1
# prev = 0
# while b <= N:
#     if f(b, N) > prev:
#         print(b, N, f(b, N))
#     prev = f(b, N)
#     b += 1
# ##########

b = 2
ans = -1
while b <= math.sqrt(N) + 1:
    if f(b, N) == S:
        ans = b
        break
    b += 1
if ans >= 0:
    print(ans)
    exit()

for i in range(1, int(math.sqrt(N)) + 1):
    hi = N // i + 1
    lo = N // (i + 1) + 1
    b = bisect_left_callable(lambda x: -f(x, N), -S, lo, hi)
    if f(b, N) == S:
        ans = b
print(ans)


Fixed submission:
import math
import os
import sys
from functools import lru_cache

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7

N, S = [int(sys.stdin.readline()) for _ in range(2)]


@lru_cache(maxsize=None)
def f(b, n):
    if n < b:
        return n
    return f(b, int(n // b)) + n % b


if S > N:
    print(-1)
    exit()
if S == N:
    print(N + 1)
    exit()

# #########
# b = 2
# ans = -1
# prev = 0
# while b <= N:
#     if f(b, N) > prev:
#         print(b, N, f(b, N))
#     prev = f(b, N)
#     b += 1
# ##########

b = 2
ans = -1
while b <= math.sqrt(N) + 1:
    if f(b, N) == S:
        ans = b
        break
    b += 1
if ans >= 0:
    print(ans)
    exit()

ans = INF
# b > sqrt(N) のとき 2 桁になるので上位桁を全探索する
for d2 in range(1, int(math.sqrt(N)) + 1):
    if (N - S) % d2 == 0:
        b = int(1 + (N - S) // d2)
        if b >= 2 and f(b, N) == S:
            ans = min(ans, b)
if ans == INF:
    print(-1)
else:
    print(ans)