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;
struct node {
  int nxt[2];
  int cnt;
  node() {
    cnt = 0;
    nxt[0] = nxt[1] = 0;
  }
};
struct trie {
  vector<node> tree;
  int n;
  trie(int n) : n(n) { tree.resize(n + 1, node()); }
  void add(long long val) {
    static int count = 0;
    int root = 0;
    for (int i = 59; i >= 0; i--) {
      int now = val >> i & 1;
      if (!tree[root].nxt[now]) {
        tree[root].nxt[now] = ++count;
      }
      root = tree[root].nxt[now];
      tree[root].cnt++;
    }
  }
  void del(long long val) {
    int root = 0;
    for (int i = 59; i >= 0; i--) {
      int now = val >> i & 1;
      int nxt = tree[root].nxt[now];
      tree[nxt].cnt--;
      if (!tree[nxt].cnt) {
        tree[root].nxt[now] = 0;
        break;
      }
      root = nxt;
    }
  }
  long long find(long long here, int mn) {
    int root = 0;
    long long res = 0;
    for (int i = 59; i >= 0; i--) {
      if (i > mn) {
        assert(tree[root].nxt[0]);
        root = tree[root].nxt[0];
      } else if (i == mn) {
        assert(tree[root].nxt[1]);
        root = tree[root].nxt[1];
        res |= 1ll << i;
      } else {
        int val = here >> i & 1;
        int now = 0;
        if (!val) {
          if (!tree[root].nxt[0]) now = 1;
        } else {
          if (tree[root].nxt[1]) now = 1;
        }
        if (now == 1) res |= 1ll << i;
        root = tree[root].nxt[now];
      }
    }
    return res;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<long long> a(n);
  vector<int> cnt(60);
  trie Tree(n * 60);
  for (auto& x : a) {
    cin >> x;
    Tree.add(x);
    for (int i = 59; i >= 0; i--) {
      if (x >> i & 1) {
        ++cnt[i];
        break;
      }
    }
  }
  long long val = 0;
  vector<long long> res;
  for (int i = 1; i <= n; i++) {
    int pos = -1;
    for (int j = 0; j < 60; j++) {
      if (val >> j & 1) continue;
      if (!cnt[j]) continue;
      pos = j;
      break;
    }
    if (pos == -1) return cout << "No", 0;
    res.emplace_back(Tree.find(val, pos));
    for (int j = 59; j >= 0; j--) {
      if (res.back() >> j & 1) {
        cnt[j]--;
        break;
      }
    }
    val ^= res.back();
    Tree.del(res.back());
  }
  cout << "Yes\n";
  for (auto& x : res) cout << x << " ";
}
