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;
template <int n>
using AI = array<int, n>;
typedef AI<2> AII;
typedef AI<3> AI3;
typedef AI<4> AI4;
const double eps = 1e-9;
const int oo = 0x3F3F3F3F;
const long long ooLL = 0x3F3F3F3F3F3F3F3FLL;
const int MOD = 1000000007;
template <typename T>
T sqr(T x) {
  return x * x;
}
const double PI = acos(-1.);
long long Round(double x) { return x < 0 ? x - .5 : x + .5; }
int main() {
  void run();
  run();
  return 0;
}
int n;
long long a[1 << 17];
void run() {
  scanf("%d", &n);
  for (int i = (0), __i(n); i < __i; ++i) scanf("%lld", &a[i]);
  vector<long long> la[64];
  for (int bi = (62), __bi(0); bi >= __bi; --bi) {
    long long b = 1LL << bi;
    for (auto &x : a)
      if (x & b) {
        la[bi].push_back(x);
        x = 0;
      }
  }
  vector<long long> rs;
  for (int bi = (62), __bi(0); bi >= __bi; --bi) {
    long long b = 1LL << bi;
    int p = 0;
    vector<long long> nx;
    for (int i = (0), __i(la[bi].size()); i < __i; ++i) {
      if (i) {
        while (p < rs.size() && (rs[p] & b) == 0) nx.push_back(rs[p++]);
        if (p == rs.size()) {
          puts("No");
          return;
        }
        nx.push_back(rs[p++]);
      }
      nx.push_back(la[bi][i]);
    }
    while (p < rs.size()) nx.push_back(rs[p++]);
    nx.swap(rs);
  }
  puts("Yes");
  for (int i = (0), __i(n); i < __i; ++i)
    printf("%lld%c", rs[i], "\n "[i + 1 < n]);
}
