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;
int n;
struct Trie {
  int trie[6000005][2];
  int cnt[6000005];
  int root = 0;
  long long val[6000005];
  long long xorsum = 0;
  long long ans[6000005 / 10];
  int C = 0;
  int newroot() {
    trie[C][0] = trie[C][1] = -1;
    cnt[C] = 0;
    return C++;
  }
  void Insert(long long x) {
    long long p = root;
    for (int i = 60; i >= 0; i--) {
      long long id = (x >> i) & 1;
      if (trie[p][id] == -1) trie[p][id] = newroot();
      p = trie[p][id];
      cnt[p]++;
    }
    val[p] = x;
  }
  bool check(long long xorsum, long long &res, int index, bool limit, int dep) {
    if (index != root && (cnt[index] == 0 || index == -1)) return false;
    if (dep == -1) {
      res = val[index];
      cnt[index]--;
      return true;
    }
    int id = (xorsum >> dep) & 1;
    if (id == 0) {
      bool ret = (check(xorsum, res, trie[index][0], limit, dep - 1)) ||
                 (!limit && check(xorsum, res, trie[index][1], limit, dep - 1));
      if (ret == 0) return false;
      cnt[index]--;
      return true;
    } else {
      bool ret = (check(xorsum, res, trie[index][0], limit, dep - 1)) ||
                 (check(xorsum, res, trie[index][1], false, dep - 1));
      if (ret == 0) return false;
      cnt[index]--;
      return true;
    }
  }
  void solve() {
    for (int i = n - 1; i >= 0; i--) {
      long long res;
      bool ret = check(xorsum, res, root, true, 60);
      if (ret == 0) {
        puts("No");
        return;
      }
      ans[i] = res;
      assert(xorsum > (xorsum ^ res));
      xorsum ^= res;
    }
    puts("Yes");
    for (int i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    puts("");
  }
} Tree;
int main() {
  scanf("%d", &n);
  Tree.root = Tree.newroot();
  for (int i = 0; i < n; i++) {
    long long x;
    scanf("%I64d", &x);
    Tree.Insert(x);
    Tree.xorsum ^= x;
  }
  Tree.solve();
}
