Given are two sequences A and B, both of length N. A and B are each sorted in the ascending order. Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it.

Constraints

* 1\leq N \leq 2 \times 10^5
* 1\leq A_i,B_i \leq N
* A and B are each sorted in the ascending order.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N


Output

If there exist no reorderings that satisfy the condition, print `No`.

If there exists a reordering that satisfies the condition, print `Yes` on the first line. After that, print a reordering of B on the second line, separating terms with a whitespace.

If there are multiple reorderings that satisfy the condition, you can print any of them.

Examples

Input

6
1 1 1 2 2 3
1 1 1 2 2 3


Output

Yes
2 2 3 1 1 1


Input

3
1 1 2
1 1 3


Output

No


Input

4
1 1 2 3
1 2 3 3


Output

Yes
3 3 1 2
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const int N = 200000;
int n, a[N], b[N], c[N + 1], d[N + 1], ans[N];

int main() {
  cin >> n;
  rep(i, n) cin >> a[i];
  rep(i, n) cin >> b[i];

  memset(c, 0, sizeof(int) * (n + 1));
  memset(c, 0, sizeof(int) * (n + 1));
  rep(i, n) c[a[i]]++;
  rep(i, n) d[b[i]]++;

  rep(i, n + 1) if (c[i] + d[i] > n) {
    cout << "No" << endl;
    return 0;
  }
  rep(i, n) c[i + 1] += c[i];
  rep(i, n) d[i + 1] += d[i];

  int x = 0;
  rep(i, n) x = max(x, c[i + 1] - d[i]);
  rep(i, n) ans[(i + x) % n] = b[i];

  cout << "Yes" << endl;
  rep(i, n - 1) cout << ans[i] << " ";
  cout << ans[n - 1] << endl;
  return 0;
}