You are given an array of $n$ integers. You need to split all integers into two groups so that the GCD of all integers in the first group is equal to one and the GCD of all integers in the second group is equal to one.

The GCD of a group of integers is the largest non-negative integer that divides all the integers in the group.

Both groups have to be non-empty.


-----Input-----

The first line contains a single integer $n$ ($2 \leq n \leq 10^5$).

The second line contains $n$ integers $a_1$, $a_2$, $\ldots$, $a_n$ ($1 \leq a_i \leq 10^9$) — the elements of the array.


-----Output-----

In the first line print "YES" (without quotes), if it is possible to split the integers into two groups as required, and "NO" (without quotes) otherwise.

If it is possible to split the integers, in the second line print $n$ integers, where the $i$-th integer is equal to $1$ if the integer $a_i$ should be in the first group, and $2$ otherwise.

If there are multiple solutions, print any.


-----Examples-----
Input
4
2 3 6 7

Output
YES
2 2 1 1 

Input
5
6 15 35 77 22

Output
YES
2 1 2 1 1 

Input
5
6 10 15 1000 75

Output
NO

Source: codeforces | Difficulty: VERY_HARD
Tags: Greedy algorithms

Number theory

Probability


Solutions:
import sys

def gcd(l):
	if len(l) == 0:
		return 0
	if len(l) == 1:
		return l[0]
	if len(l) == 2:
		if l[1] == 0:
			return l[0]
		return gcd([l[1], l[0] % l[1]])
	return gcd([gcd(l[:-1]), l[-1]])

def brute_force(l1, l2, l, sol):
	if len(l) == 0:
		g1 = gcd(l1)
		g2 = gcd(l2)
		return (g1 == 1 and g2 == 1, sol)
	(res, s) = brute_force(l1 + [l[0]], l2, l[1:], sol + [1])
	if res:
		return (True, s)
	return brute_force(l1, l2 + [l[0]], l[1:], sol + [2])

def factor(n):
	res = []
	i = 2
	while i * i <= n:
		if n % i == 0:
			res.append(i)
		while n % i == 0:
			n = int(n / i)
		i += 1
	if n != 1:
		res.append(n)
	return res

def dumpsol(sol):
	for v in sol:
		print(v, end=' ')
n = int(sys.stdin.readline())
l = sys.stdin.readline().strip().split(' ')[0:n]
l = [int(x) for x in l]
if n < 12:
	(ret, sol) = brute_force([], [], l, [])
	if ret:
		print('YES')
		dumpsol(sol)
	else:
		print('NO')
	return
factors = {}
for i in range(10):
	for key in factor(l[i]):
		factors[key] = 0
flists = {}
for f in factors:
	flists[f] = []
	pos = 0
	found = False
	for v in l:
		if v % f != 0:
			found = True
			factors[f] += 1
			flists[f].append(pos)
			if factors[f] > 9:
				break
		pos += 1
	if not found:
		print('NO')
		return
oftf = []
isoftf = {}
for f in factors:
	if factors[f] == 0:
		print('NO')
		return
	if factors[f] < 10:
		oftf.append(f)
		isoftf[f] = 1
sol = [1 for i in range(len(l))]
x = l[0]
sol[0] = 2
oxf = factor(x)
xf = []
nxf = 0
isxoftf = {}
for f in oxf:
	if f in isoftf:
		nxf += 1
		isxoftf[f] = 1
		xf.append(f)
	else:
		sol[flists[f][0]] = 2
nonxf = []
for f in oftf:
	if not f in isxoftf:
		nonxf.append(f)
masks = {}
pos = 0
for f in xf + nonxf:
	for v in flists[f]:
		if not v in masks:
			masks[v] = 0
		masks[v] |= 1 << pos
	pos += 1
vals = [{} for i in range(len(masks) + 1)]
vals[0][0] = 0
pos = 0
mlist = []
for mask in masks:
	mlist.append(mask)
	cmask = masks[mask]
	cmask1 = cmask << 10
	for v in vals[pos]:
		vals[pos + 1][v | cmask] = v
		if mask != 0:
			vals[pos + 1][v | cmask1] = v
	pos += 1
test_val = (1 << len(xf)) - 1 | (1 << len(oftf)) - 1 << 10
for v in vals[pos]:
	if v & test_val == test_val:
		print('YES')
		while pos != 0:
			nv = vals[pos][v]
			if nv ^ v < 1024 and nv ^ v != 0:
				sol[mlist[pos - 1]] = 2
			v = nv
			pos -= 1
		dumpsol(sol)
		return
print('NO')