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.
#include <bits/stdc++.h>
using namespace std;
int n;
long long b[62][100005];
long long a[100005], tot;
long long c[62];
long long ans[100005];
bool pd;
void init() { freopen("e.in", "r", stdin); }
int main() {
  c[1] = 1;
  for (int i = 2; i <= 60; i++) c[i] = c[i - 1] * 2;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    for (int j = 60; j >= 1; j--)
      if (a[i] >= c[j]) {
        b[j][++b[j][0]] = a[i];
        break;
      }
  }
  tot = 0;
  for (int i = 1; i <= n; i++) {
    pd = 0;
    for (int j = 1; j <= 60; j++) {
      if (tot / c[j] % 2 == 0) {
        if (b[j][0]) {
          tot = tot xor b[j][b[j][0]];
          ans[i] = b[j][b[j][0]];
          b[j][0]--;
          pd = 1;
          break;
        }
      }
    }
    if (!pd) {
      printf("No");
      return 0;
    }
  }
  printf("Yes\n");
  for (int i = 1; i <= n; i++) printf("%I64d ", ans[i]);
  return 0;
}
