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;
const int N = 666666;
const int M = 666;
const long long md = 1e9 + 7;
long long addmod(long long a, long long b) { return (a + b) % md; }
template <class T>
void mine(T &a, T b) {
  if (a > b) a = b;
}
int n, m, k, q;
int ans;
long long a[N];
int pre[N], nxt[N];
int c[N];
vector<int> v[66], ins;
void print() {
  for (int i = nxt[0]; i; i = nxt[i])
    printf("%I64d%c", a[i], nxt[i] ? ' ' : '\n');
}
void add(int m, int p) {
  pre[nxt[p]] = m;
  nxt[m] = nxt[p];
  pre[m] = p;
  nxt[p] = m;
}
int fuck() {
  for (int i = 60; i >= 0; i--)
    if (v[i].size()) {
      if (v[i].size() > 1) return 0;
      m = i;
      add(v[i][0], 0);
      break;
    }
  for (int i = m - 1; i >= 0; i--) {
    if (v[i].size()) {
      ins.clear();
      ins.push_back(0);
      for (int j = 1, cur = nxt[0]; j < v[i].size(); j++, cur = nxt[cur]) {
        while (cur && !(a[cur] & (1ll << i))) cur = nxt[cur];
        if (j && !cur) return 0;
        ins.push_back(cur);
      }
      for (int j = 0; j < v[i].size(); j++) add(v[i][j], ins[j]);
    }
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(false);
  cout << setprecision(20);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 60; j >= 0; j--)
      if (a[i] & (1ll << j)) {
        v[j].push_back(i);
        break;
      }
  }
  ans = fuck();
  if (ans) {
    puts("Yes");
    print();
  } else
    puts("No");
  return 0;
}
