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;
using ll = long long;
const int maxn = 1e5 + 100;
ll b[maxn];
int trie[maxn * 50][2], tot, cnt[maxn * 50];
void insert(ll x) {
  int u = 0;
  for (int i = 61 - 1; i >= 0; --i) {
    int tp = (x >> i) & 1;
    if (!trie[u][tp]) trie[u][tp] = ++tot;
    u = trie[u][tp];
    cnt[u]++;
  }
}
void del(ll x) {
  int u = 0;
  for (int i = 61 - 1; i >= 0; --i) {
    int tp = (x >> i) & 1;
    int k = u;
    u = trie[u][tp];
    cnt[u]--;
    if (!cnt[u]) trie[k][tp] = 0;
  }
}
ll find(ll x, int rt, int p, bool limt, ll k) {
  if (p == -1) return limt ? -2e18 : 0;
  ll ws = ((x >> p) & 1), ret = 0;
  ll top = limt ? ((k >> p) & 1) : 1;
  if (trie[rt][!ws] && (top)) {
    ret = 1LL << p;
    ret += find(x, trie[rt][!ws], p - 1, limt, k);
    if (ret >= 0) return ret;
  }
  if (trie[rt][ws]) {
    ret = find(x, trie[rt][ws], p - 1, limt && top == 0, k);
    if (ret >= 0) return ret;
  }
  return -2e18;
}
ll a[maxn];
int main() {
  int n;
  scanf("%d", &n);
  ll sum = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &b[i]);
    insert(b[i]);
    sum ^= b[i];
  }
  int flag = 0;
  a[n] = sum;
  a[0] = 0;
  for (int i = n - 1; i >= 1; --i) {
    ll tp = find(sum, 0, 61 - 1, 1, sum);
    if (tp < sum && tp >= 0)
      a[i] = tp;
    else {
      flag = 1;
      break;
    }
    del(sum ^ tp);
    sum = tp;
  }
  if (flag)
    puts("No");
  else {
    puts("Yes");
    for (int i = 1; i <= n; ++i) printf("%lld ", a[i] ^ a[i - 1]);
  }
  return 0;
}
