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 double PI = 3.14159265358979323846;
const double eps = (1e-5);
int isGreater(long double x, long double y) {
  if (abs(x - y) < eps) return 0;
  if (x > y) return 1;
  return -1;
}
int n;
vector<long long> b;
vector<vector<int> > v;
vector<int> isEnd;
vector<int> visitCnt;
int addNode() {
  v.push_back(vector<int>(2, -1));
  isEnd.push_back(0);
  visitCnt.push_back(0);
  return (int)(isEnd).size() - 1;
}
const long long INF = (1LL << 62);
void insert(long long x, int u = 0, int i = 60) {
  visitCnt[u]++;
  if (i == -1) {
    isEnd[u]++;
    return;
  }
  int to = (x >> i) & 1;
  if (v[u][to] == -1) v[u][to] = addNode();
  insert(x, v[u][to], i - 1);
}
long long find(long long x, long long soFar, int must, int u = 0, int i = 60) {
  if (i == -1) return soFar;
  int to = (x >> i) & 1;
  if (v[u][to] != -1)
    return find(x, soFar | ((1LL << i) * to), must, v[u][to], i - 1);
  else if (v[u][!to] != -1) {
    if (i >= must) return INF;
    return find(x, soFar | ((1LL << i) * (!to)), must, v[u][!to], i - 1);
  }
  return INF;
}
bool erase(long long x, int u = 0, int i = 60) {
  visitCnt[u]--;
  if (i == -1) {
    isEnd[u]--;
    return visitCnt[u] == 0;
  }
  int to = (x >> i) & 1;
  assert(v[u][to] != -1);
  bool zero = erase(x, v[u][to], i - 1);
  if (zero) v[u][to] = -1;
  return visitCnt[u] == 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  b.resize(n);
  addNode();
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
    insert(b[i]);
  }
  long long x = 0;
  vector<long long> out(n);
  for (int cnt = 0; cnt < n; ++cnt) {
    long long minXor = INF, minY;
    for (int i = 0; i <= 60; ++i) {
      if ((x >> i) & 1) continue;
      long long f = (1LL << i);
      for (int j = i - 1; j >= 0; --j) {
        if ((1LL << j) & x) f |= (1LL << j);
      }
      long long y = find(f, 0, i);
      if (y == INF) continue;
      long long tempXor = (y ^ x);
      if (tempXor < minXor) {
        minXor = tempXor;
        minY = y;
      }
    }
    if (minXor == INF) {
      cout << "No\n";
      return 0;
    }
    x = minXor;
    out[cnt] = minY;
    erase(minY);
  }
  cout << "Yes\n";
  for (long long item : out) cout << item << ' ';
  return 0;
}
