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>
long long bin[100], num[1000000], hd[1000000], en[1000000], ans[1000000];
long long bsf(long long a) {
  long long i, j;
  for (i = 1, j = 1; i <= 62; i++, j = j << 1) {
    if (a & 1) {
      return j - 1;
    }
  }
  return 0;
}
int main() {
  long long i, j, k, a, b, c, n, m, t;
  scanf("%lld", &n);
  bin[1] = 1;
  for (i = 2; i <= 62; i++) {
    bin[i] = bin[i - 1] << 1;
  }
  for (i = 1; i <= n; i++) {
    scanf("%lld", &num[i]);
  }
  std::sort(num + 1, num + n + 1);
  memset(hd, 0, sizeof(*hd) * (100));
  memset(en, 0, sizeof(*en) * (100));
  j = 1;
  k = 0;
  for (i = 1; i <= n; i++) {
    while (j <= num[i]) {
      j = j << 1;
      k += 1;
    }
    if (hd[k] == 0) {
      hd[k] = i;
    }
  }
  en[k] = n;
  a = hd[k];
  for (i = k - 1; i >= 1; i--) {
    if (hd[i] > 0) {
      en[i] = a - 1;
      a = hd[i];
    }
  }
  a = 0;
  for (i = 1; i <= n; i++) {
    a = a ^ num[i];
  }
  *ans = 0;
  t = 0;
  while (a > 0) {
    for (i = 1, j = 1; i <= 62; i++, j = j << 1) {
      if (a & j) {
        if ((hd[i] > 0) && (hd[i] <= en[i])) {
          ans[++(*ans)] = num[hd[i]];
          a = a ^ num[hd[i]];
          hd[i] += 1;
          break;
        }
      }
    }
    if (i > 62) {
      goto FAIL;
    }
  }
  for (i = 1; i <= 62; i++) {
    if ((hd[i] > 0) && (hd[i] <= en[i])) {
      goto FAIL;
    }
  }
  printf("Yes\n");
  for (i = *ans; i >= 1; i--) {
    printf("%lld ", ans[i]);
  }
  printf("\n");
  return 0;
FAIL:
  printf("No\n");
  return 0;
}
