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:16777216")
using namespace std;
struct Timer {
  static map<string, Timer> ar;
  long long time;
  long long last;
  bool running;
  int count;
  Timer() : time(0), running(false), count(0) {}
  void start() {}
  void stop() {}
  static void start(string name) {}
  static void stop(string name) {}
  long double time_f() { return 0; }
  long double avg_time_f() { return 0; }
};
map<string, Timer> Timer::ar;
struct TimerRegion {
  string name;
  TimerRegion(string _name) : name(_name) { Timer::start(name); }
  ~TimerRegion() { Timer::stop(name); }
};
const int MAXN = (int)1e5 + 10;
long long ar[MAXN];
vector<int> by_bit[60];
long long ans[MAXN];
long long ansl;
long long doit(int bit, long long cur) {
  if (bit != 0) cur = doit(bit - 1, cur);
  while (!by_bit[bit].empty() && (cur ^ ar[by_bit[bit].back()]) > cur) {
    cur ^= ar[by_bit[bit].back()];
    ans[ansl++] = by_bit[bit].back();
    by_bit[bit].pop_back();
    if (bit != 0) cur = doit(bit - 1, cur);
  }
  return cur;
}
void solve() {
  TimerRegion timer_func(__FUNCTION__);
  ;
  ansl = 0;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", &ar[i]);
  sort(ar, ar + n);
  reverse(ar, ar + n);
  int highest_bit = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 60; ++j)
      if (ar[i] >= (1ll << j) && ar[i] < (1ll << (j + 1))) {
        by_bit[j].push_back(i);
        highest_bit = max(highest_bit, j);
        break;
      }
  doit(highest_bit, 0);
  if (ansl == n) {
    printf("Yes\n");
    for (int i = 0; i < n; ++i) printf("%lld ", ar[ans[i]]);
  } else
    printf("No\n");
}
int main() {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}
