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 = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 62;
int getBit(long long a) {
  int res = 0;
  while (a) {
    a /= 2;
    res++;
  }
  return res - 1;
}
vector<long long> A[MAX];
vector<long long> solve(vector<long long>& a) {
  for (int i = (0); i < (MAX); i++) {
    A[i].clear();
  }
  for (int i = (0); i < ((int)a.size()); i++) {
    int b = getBit(a[i]);
    A[b].push_back(a[i]);
  }
  long long cur = 0;
  vector<long long> res;
  for (int it = (0); it < ((int)a.size()); it++) {
    int ind = -1;
    for (int i = (0); i < (MAX); i++) {
      if (cur & (1ll << i)) continue;
      if (A[i].empty()) continue;
      ind = i;
      break;
    }
    if (ind == -1) return vector<long long>();
    cur ^= A[ind].back();
    res.push_back(A[ind].back());
    A[ind].pop_back();
  }
  return res;
}
bool check(vector<long long> a) {
  int cur = 0;
  for (int i = (0); i < ((int)a.size()); i++) {
    if ((cur ^ a[i]) <= cur) return false;
    cur ^= a[i];
  }
  return true;
}
void print(vector<long long> a) {
  cout << (int)a.size() << ": ";
  for (int i = (0); i < ((int)a.size()); i++) {
    cout << a[i] << ' ';
  }
  cout << endl;
}
vector<long long> gen(int n) {
  while (true) {
    vector<long long> a;
    for (int i = (0); i < (n); i++) {
      a.push_back(rand() % 1023 + 1);
    }
    sort(a.begin(), a.end());
    a.erase(unique(a.begin(), a.end()), a.end());
    vector<long long> b;
    b.push_back(a[0]);
    bool ok = true;
    for (int i = (1); i < ((int)a.size()); i++) {
      b.push_back(a[i] ^ a[i - 1]);
      if (b.back() == 0) {
        ok = false;
        break;
      }
    }
    if (!ok) continue;
    return b;
  }
}
int main() {
  int n;
  vector<long long> a;
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) {
    long long x;
    scanf("%I64d", &x);
    a.push_back(x);
  }
  vector<long long> res = solve(a);
  if (res.empty())
    printf("No\n");
  else {
    printf("Yes\n");
    for (int i = (0); i < ((int)res.size()); i++) {
      printf("%I64d ", res[i]);
    }
  }
  printf("\n");
}
