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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\n";
  err(++it, args...);
}
const double PI = acos(-1);
const double eps = (1e-9);
const long long int INF = 2e18;
const int M = (1e9) + 7;
const int fft = 998244353;
const int N = (4e5) + 3;
const int B = 62;
vector<long long int> ad[B];
int cnt[B], num[B];
int bit(long long int y) {
  int a = 0;
  while (y > 1) a++, y /= 2;
  return a;
}
void solve() {
  int n, i, b;
  scanf("%d", &n);
  long long int a, x = INF;
  for (i = 0; i < n; ++i)
    scanf("%lld", &a), b = bit(a), ad[b].push_back(a), x = min(x, a);
  for (i = 0; i < B; ++i)
    sort(ad[i].begin(), ad[i].end()), num[i] = (int)ad[i].size();
  vector<long long int> ans;
  ans.push_back(x);
  b = bit(x);
  cnt[b]++;
  while (1) {
    for (i = 0; i < B; ++i) {
      if ((!((x >> i) & 1)) && cnt[i] < num[i]) break;
    }
    if (i == B) break;
    a = ad[i][cnt[i]];
    ans.push_back(a);
    x ^= a, cnt[i]++;
  }
  if ((int)ans.size() < n)
    printf("No\n");
  else {
    printf("Yes\n");
    for (i = 0; i < n; ++i) cout << ans[i] << " ";
  }
}
int main() {
  int t = 1;
  while (t--) solve();
}
