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;
long long sqr(long long x) { return x * x; }
int mysqrt(long long x) {
  int l = 0, r = 1e9 + 1;
  while (r - l > 1) {
    int m = (l + r) / 2;
    if (m * (long long)m <= x)
      l = m;
    else
      r = m;
  }
  return l;
}
mt19937 rnd(1227);
mt19937_64 rndll(12365);
long long AR = 19, BR = 13, CR = 23, XR = 228, YR = 322, MODR = 1e9 + 993;
long long myrand() {
  long long ZR = (XR * AR + YR * BR + CR) % MODR;
  XR = YR;
  YR = ZR;
  return ZR;
}
const int Mod = 1e9 + 7;
int bpow(int x, int y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  int ret = bpow(x, y >> 1);
  ret = (ret * (long long)ret) % Mod;
  if (y & 1) ret = (ret * (long long)x) % Mod;
  return ret;
}
int bdiv(int x, int y) { return (x * (long long)bpow(y, Mod - 2)) % Mod; }
void setmin(int &x, int y) { x = min(x, y); }
void setmax(int &x, int y) { x = max(x, y); }
void setmin(long long &x, long long y) { x = min(x, y); }
void setmax(long long &x, long long y) { x = max(x, y); }
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
const long long llinf = 2e18 + 100;
const double eps = 1e-9;
const int maxn = 3e5 + 100, maxw = 1e6 + 100, inf = 2e9 + 100, sq = 300,
          mod = 1e9 + 993, LG = 17;
int n;
long long a[maxn];
long long mas[maxn], arr[maxn];
bool f(long long x, int i) { return (x & (1ll << i)) > 0; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  reverse(a, a + n);
  int it = 0;
  int cnt = 1;
  for (int i = 59; i >= 0; i--) {
    int last = it;
    while (last < n && f(a[last], i)) last++;
    int t = 0;
    int nt = 0;
    for (int j = 0; j < cnt; j++) {
      arr[nt++] = mas[j];
      if (f(mas[j], i)) t ^= 1;
      if (t == 0 && it < last) arr[nt++] = a[it], it++, t ^= 1;
    }
    for (int j = 0; j < nt; j++) mas[j] = arr[j];
    cnt = nt;
    if (it != last) {
      cout << "No";
      return 0;
    }
  }
  cout << "Yes\n";
  for (int i = 1; i <= n; i++) cout << mas[i] << ' ';
}
