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;
inline long long GCD(long long a, long long b) {
  while (b != 0) {
    int t = a;
    a = b;
    b = t % b;
  }
  return a;
}
inline long long LCM(long long a, long long b) {
  if (a == 0 || b == 0) return GCD(a, b);
  return a * b / GCD(a, b);
}
inline long long CEIL(long long n, long long d) { return n / d + n % d != 0; }
inline long long ROUND(long long n, long long d) {
  return n / d + (long long)((n % d) * 2 >= d);
}
inline long long POW(long long a, long long n) {
  if (n < 0) return 0;
  long long ret = 1;
  while (n) {
    if (n % 2) ret *= a;
    a = a * a;
    n /= 2;
  }
  return ret;
}
const int MAXN = 1e5;
int n;
long long ar[MAXN];
vector<long long> s[64];
vector<long long> res, t;
void input() {
  cin >> n;
  for (int(i) = (0); (i) < (int)(n); ++(i)) cin >> ar[i];
}
int solve() {
  for (int(i) = (0); (i) < (int)(n); ++(i))
    s[63 - __builtin_clzll(ar[i])].push_back(ar[i]);
  for (int(i) = (0); (i) < (int)(64); ++(i))
    sort(s[i].begin(), s[i].end(),
         [](long long u, long long v) { return u > v; });
  for (int(i) = (63); (i) >= (int)(0); --(i)) {
    if (s[i].size() == 0) continue;
    int cnt = 0;
    for (int(j) = (0); (j) < (int)(res.size()); ++(j))
      if (res[j] & (1ll << i)) ++cnt;
    if (s[i].size() > cnt + 1) return printf("No\n");
    t.clear();
    int x = 0, y = 0, c = 0;
    while (x < s[i].size() || y < res.size()) {
      if (x == c && x < s[i].size())
        t.push_back(s[i][x++]);
      else {
        if (res[y] & (1ll << i)) ++c;
        t.push_back(res[y++]);
      }
    }
    res = t;
  }
  assert(res.size() == n);
  printf("Yes\n");
  for (int(i) = (0); (i) < (int)(n); ++(i)) printf("%lld ", res[i]);
  return printf("\n");
}
void execute() { input(), solve(); }
int main(void) {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  execute();
}
