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;
const long long N = 1e5;
const long long M = 1e4;
const long long mod = 1e9 + 7;
const long long inf = 1e9;
long long read() {
  long long s = 0;
  register bool neg = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar()) neg |= (c == '-');
  for (; c >= '0' && c <= '9'; s = s * 10 + (c ^ 48), c = getchar())
    ;
  s = (neg ? -s : s);
  return s;
}
long long a, s[N + 5], r[N + 5];
long double val[N + 5];
vector<long long> p[63], q[63];
bool cmp(long long p1, long long p2) { return val[p1] < val[p2]; }
signed main() {
  a = read();
  for (long long i = (1); i <= (a); ++i) {
    s[i] = read();
    long long j = 0, x = s[i];
    for (; x; ++j)
      if (x & (1ll << j)) {
        if ((x ^ (1ll << j))) q[j].push_back(i);
        x ^= (1ll << j);
      }
    --j;
    p[j].push_back(i);
  }
  long long L = 0;
  for (long long i = (61); i >= (0); --i) {
    if (!((long long)((p[i]).size()))) continue;
    if (((long long)((p[i]).size())) > ((long long)((q[i]).size())) + 1) {
      puts("No");
      return 0;
    }
    sort(q[i].begin(), q[i].end(), cmp);
    r[p[i][0]] = L;
    val[p[i][0]] = val[L] - 1e18;
    L = p[i][0];
    for (long long j = (1); j <= (((long long)((p[i]).size())) - 1); ++j) {
      long long k = p[i][j], t = q[i][j - 1];
      val[k] = (val[t] + val[r[t]]) / 2;
      r[k] = r[t];
      r[t] = k;
    }
  }
  puts("Yes");
  for (long long i = L; i; i = r[i]) printf("%lld ", s[i]);
  return 0;
}
