Vitya has learned that the answer for The Ultimate Question of Life, the Universe, and Everything is not the integer 54 42, but an increasing integer sequence a_1, …, a_n. In order to not reveal the secret earlier than needed, Vitya encrypted the answer and obtained the sequence b_1, …, b_n using the following rules:

  * b_1 = a_1;
  * b_i = a_i ⊕ a_{i - 1} for all i from 2 to n, where x ⊕ y is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of x and y. 



It is easy to see that the original sequence can be obtained using the rule a_i = b_1 ⊕ … ⊕ b_i.

However, some time later Vitya discovered that the integers b_i in the cypher got shuffled, and it can happen that when decrypted using the rule mentioned above, it can produce a sequence that is not increasing. In order to save his reputation in the scientific community, Vasya decided to find some permutation of integers b_i so that the sequence a_i = b_1 ⊕ … ⊕ b_i is strictly increasing. Help him find such a permutation or determine that it is impossible.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5).

The second line contains n integers b_1, …, b_n (1 ≤ b_i < 2^{60}).

Output

If there are no valid permutations, print a single line containing "No".

Otherwise in the first line print the word "Yes", and in the second line print integers b'_1, …, b'_n — a valid permutation of integers b_i. The unordered multisets \\{b_1, …, b_n\} and \\{b'_1, …, b'_n\} should be equal, i. e. for each integer x the number of occurrences of x in the first multiset should be equal to the number of occurrences of x in the second multiset. Apart from this, the sequence a_i = b'_1 ⊕ … ⊕ b'_i should be strictly increasing.

If there are multiple answers, print any of them.

Examples

Input

3
1 2 3


Output

No


Input

6
4 7 7 12 31 61


Output

Yes
4 12 7 31 7 61 

Note

In the first example no permutation is valid.

In the second example the given answer lead to the sequence a_1 = 4, a_2 = 8, a_3 = 15, a_4 = 16, a_5 = 23, a_6 = 42.
from bisect import bisect
import atexit
import io
import sys

_INPUT_LINES = sys.stdin.read().splitlines()
raw_input = iter(_INPUT_LINES).next
_OUTPUT_BUFFER = io.BytesIO()
sys.stdout = _OUTPUT_BUFFER

@atexit.register
def write():
    sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())

def interleave(c, b, mask):
	lb, ib, s = len(b), 0, False
	if ib<lb:
		yield b[ib]
		ib, s = 1, True
	for e in c:
		yield e
		s ^= (e&mask)>0
		if not s and ib<lb:
			yield b[ib]
			ib, s = ib+1, True
	if ib<lb: yield -1

def calprefix(b):
	cur = 0
	for e in b:
		cur ^= e
		yield cur
	
n = int(raw_input())
b = [int(t) for t in raw_input().split()]
c = []
for w in range(60,0,-1):
	low, high = 1<<(w-1), (1<<w)-1
	bx = filter(lambda x: low<=x and x<=high, b)
	c = list(interleave(c, bx, low))
	if len(c)>0 and c[-1]==-1:
		print "No"
		sys.exit(0)
print "Yes"
print " ".join(map(lambda x: str(x), c))
