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<iostream>
#include <bits/stdc++.h>
#define ll long long
#define ld  long double
#define IO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int N=2e5+2;
int n, a[N], b[N];
vector<int> v;
int main()
{
    IO
    cin>>n;
    for(int i=0; i<n; i++)
        cin>>a[i];
    for(int i=0; i<n; i++)
        cin>>b[i];
    sort(b,b+n,greater<int>());
    for(int i=0; i<n; i++)
        if(a[i]==b[i])
            v.push_back(i);
    int j=0;
    for(auto i : v)
    {
        while(j<n && (a[j]==b[i] || b[j]==a[i]))
            j++;
        if(j==n)
            return cout<<"No", 0;
        swap(b[j],b[i]);
    }
    cout<<"Yes\n";
    for(int i=0; i<n; i++)
        cout<<b[i]<<" ";
}
