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;
mt19937 gen(chrono::system_clock::now().time_since_epoch().count());
const int M = 1e9 + 7;
const int N = 7e6 + 7;
long long n, cur = 3e18, V, ans[N], x;
long long nxt[N][2], k[N];
void add(long long x) {
  int v = 0;
  for (long long i = 60; i >= 0; i--) {
    long long to = (x >> i) & 1ll;
    if (!nxt[v][to]) nxt[v][to] = ++V;
    v = nxt[v][to];
    k[v]++;
  }
  return;
}
void del(long long x) {
  int v = 0;
  for (long long i = 60; i >= 0; i--) {
    long long to = (x >> i) & 1ll;
    k[nxt[v][to]]--;
    int vv = nxt[v][to];
    if (!k[nxt[v][to]]) nxt[v][to] = 0;
    v = vv;
  }
  return;
}
long long mn(long long x, int p, int u) {
  int v = u;
  long long res = 0;
  for (long long i = p; i >= 0; i--) {
    long long to = ((x >> i) & 1ll);
    if (nxt[v][to])
      v = nxt[v][to];
    else {
      v = nxt[v][to ^ 1];
      res += 1ll << i;
    }
  }
  return res;
}
long long q(long long x) {
  int v = 0;
  long long ans = 0;
  long long res = 0;
  for (long long i = 60; i >= 0; i--) {
    long long to = ((x >> i) & 1ll);
    if (nxt[v][to ^ 1]) {
      res += 1ll << i;
      v = nxt[v][to ^ 1];
    } else
      v = nxt[v][to];
  }
  ans = res;
  res = 0;
  v = 0;
  for (long long i = 60; i >= 0; i--) {
    long long to = ((x >> i) & 1);
    if (to == 1) {
      if (nxt[v][to ^ 1]) {
        res += 1ll << i;
        v = nxt[v][to ^ 1];
      } else
        break;
    } else {
      if (nxt[v][to ^ 1]) {
        long long curr = res + (1ll << i) + mn(x, i - 1, nxt[v][to ^ 1]);
        if (curr > x) ans = min(ans, curr);
      }
      if (nxt[v][to])
        v = nxt[v][to];
      else
        break;
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x;
    add(x);
    cur = min(cur, x);
  }
  ans[0] = cur;
  del(cur);
  for (int i = 1; i < n; i++) {
    long long now = q(cur);
    del(now ^ cur);
    if (now <= cur) {
      cout << "No";
      return 0;
    }
    ans[i] = cur ^ now;
    cur = now;
  }
  cout << "Yes" << '\n';
  for (int i = 0; i < n; i++) cout << ans[i] << ' ';
  return 0;
}
