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 bad() {
  cout << "No\n";
  exit(0);
}
const int N = 1e6;
int n;
long long b[N];
bool isbit(long long mask, int bit) { return ((1ll << bit) & mask) != 0; }
int sz[N];
vector<vector<int> > trie[60];
long long get(int st, long long xr) {
  int vr = 0;
  long long res = 0;
  for (int j = 59; j >= 0; j--) {
    int bt = isbit(xr, j);
    if (trie[st][trie[st][vr][bt]][2] != 0) {
      res += (1ll << j) * bt;
      trie[st][trie[st][vr][bt]][2]--;
      vr = trie[st][vr][bt];
    } else {
      bt = 1 - bt;
      res += (1ll << j) * bt;
      trie[st][trie[st][vr][bt]][2]--;
      vr = trie[st][vr][bt];
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  vector<int> nill = {0, 0, 0};
  for (int i = 0; i < 60; i++) trie[i].push_back(nill);
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    int vr = 0, st = 0;
    for (int j = 59; j >= 0; j--)
      if (isbit(b[i], j)) {
        st = j;
        break;
      }
    sz[st]++;
    for (int j = 59; j >= 0; j--) {
      int bt = isbit(b[i], j);
      if (trie[st][vr][bt] == 0) {
        trie[st].push_back(nill);
        trie[st][vr][bt] = trie[st].size() - 1;
      }
      vr = trie[st][vr][bt];
      trie[st][vr][2]++;
    }
  }
  long long xr = 0;
  vector<long long> ans;
  for (int i = 1; i <= n; i++) {
    int found = 0;
    for (int j = 0; j < 60; j++)
      if (!isbit(xr, j) && sz[j] > 0) {
        found = 1;
        sz[j]--;
        long long q = get(j, xr);
        ans.push_back(q);
        xr ^= q;
        break;
      }
    if (!found) bad();
  }
  cout << "Yes\n";
  for (int i = 1; i <= n; i++) cout << ans[i - 1] << " ";
  cout << "\n";
  return 0;
}
