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>
#pragma warning(disable : 4996)
using namespace std;
template <class T>
inline void chkmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void chkmax(T &a, T b) {
  if (a < b) a = b;
}
const int MAXN = 1e5 + 7;
const int MAXL = 60;
long long ans[2][MAXN];
long long b[MAXN];
bool Less(long long x, long long y) { return x > y; }
bool GetBit(long long n, long long i) { return (n >> i) & 1ll; }
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
  }
  sort(b, b + n, Less);
  int i = 0, j = 0;
  bool q = 0;
  int sz = 0;
  for (int l = MAXL - 1; l >= 0; --l) {
    while (j < n && GetBit(b[j], l)) ++j;
    bool w = !q;
    int kk = 0, k = 0;
    while (kk < sz) {
      if (GetBit(ans[q][kk], l) && i < j) {
        ans[w][k++] = b[i++];
      }
      ans[w][k++] = ans[q][kk++];
    }
    if (i < j) {
      ans[w][k++] = b[i++];
      if (i < j) {
        cout << "No\n";
        return 0;
      }
    }
    sz = k;
    q = w;
  }
  cout << "Yes\n";
  for (int i = 0; i < n; ++i) {
    cout << ans[q][i] << ' ';
  }
  cout << '\n';
}
