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 comment(linker, "/STACK:1000000000")
using namespace std;
const int maxn = (int)1e5 + 10;
long long v[maxn];
bool used[maxn];
int g[maxn];
int main() {
  int n;
  cin >> n;
  int pre = 0;
  for (int i = 0; i < n; i++) {
    g[i] = pre + rand() % 15;
    pre = g[i];
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &v[i]);
  }
  vector<long long> f;
  for (int b = 59; b >= 0; b--) {
    vector<long long> cur;
    for (int i = 0; i < n; i++) {
      if (!used[i] && ((v[i] >> b) & 1)) {
        used[i] = true;
        cur.push_back(v[i]);
      }
    }
    vector<long long> now;
    int pos = 0;
    for (int i = 0; i < (int)cur.size(); i++) {
      while (pos != (int)f.size() && ((f[pos] >> b) & 1) == 0) {
        now.push_back(f[pos]);
        pos++;
      }
      if (pos == (int)f.size()) {
        if (i + 1 == (int)cur.size()) {
          now.push_back(cur[i]);
        } else {
          printf("No\n");
          return 0;
        }
      } else {
        now.push_back(cur[i]);
        now.push_back(f[pos++]);
      }
    }
    while (pos != (int)f.size()) {
      now.push_back(f[pos++]);
    }
    f = now;
  }
  printf("Yes\n");
  long long last = 0;
  for (int i = 0; i < (int)f.size(); i++) {
    if (last > (last ^ f[i])) {
      throw 1;
    }
    last ^= f[i];
    printf("%lld ", f[i]);
  }
  printf("\n");
  return 0;
}
