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 <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  for (auto &vi : vec) os << vi << " ";
  return os;
}
template <typename T, int B>
struct BinaryTrie {
  struct Node {
    int cnt;
    Node *ch[2];
    Node() : cnt(0), ch{nullptr, nullptr} {}
  };
  Node *root;
  BinaryTrie() { root = new Node(); }
  int size() { return root->cnt; }
  inline int size(Node *x) { return x ? x->cnt : 0; }
  inline void add(T x, int c) {
    Node *cur = root;
    cur->cnt += c;
    for (int b = B - 1; b >= 0; b--) {
      bool f = x >> b & 1;
      if (!cur->ch[f]) cur->ch[f] = new Node();
      cur = cur->ch[f];
      cur->cnt += c;
    }
  }
  void insert(T x) { add(x, 1); }
  void erase(T x) { add(x, -1); }
  inline T get(int k, T y = 0) {
    Node *cur = root;
    T res = 0;
    for (int b = B - 1; b >= 0; b--) {
      bool g = y >> b & 1;
      int m = size(cur->ch[g]);
      if (k >= m) k -= m, g ^= 1;
      cur = cur->ch[g], res |= T(g) << b;
    }
    return res;
  }
  T min_element(T y = 0) { return get(0, y); }
  T max_element(T y = 0) { return get(size() - 1, y); }
  inline int count_less(T x, T y = 0) {
    Node *cur = root;
    int res = 0;
    for (int b = B - 1; b >= 0; b--) {
      bool f = x >> b & 1;
      bool g = y >> b & 1;
      if (f) res += size(cur->ch[g]);
      cur = cur->ch[f ^ g];
      if (size(cur) == 0) return res;
    }
    return res;
  }
  T lower_bound(T x, T y = 0) { return get(count_less(x, y), y); }
  T upper_bound(T x, T y = 0) { return get(count_less(x + 1, y), y); }
};
int main() {
  int n;
  cin >> n;
  BinaryTrie<long long, 60> bt;
  for (int i = 0; i < n; i++) {
    long long b;
    cin >> b;
    bt.insert(b);
  }
  long long x = 0;
  vector<long long> res(n);
  for (int i = 0; i < n; i++) {
    if ((bt.max_element(x) ^ x) <= x) return !(cout << "No" << endl);
    long long val = bt.upper_bound(x, x);
    bt.erase(val);
    x ^= res[i] = val;
  }
  cout << "Yes" << endl;
  cout << res << endl;
  return 0;
}
