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>
using namespace std;
typedef long long ll;
#define rep(i,a,b) for (ll i = (a); i < (b); i++)
#define REP(i,n) rep(i,0,n)
#define mod (1000000007)

void solve() 
{
    int n;cin>>n;
    int a[n],b[n],c[n+1],d[n+1];
    REP(i,n)cin>>a[i];REP(i,n)cin>>b[i];
    REP(i,n+1)c[i]=d[i]=0;
    REP(i,n)c[a[i]]++,d[b[i]]++;
    REP(i,n+1)if(c[i]+d[i]>n){cout<<"No"<<endl;return;}
    cout<<"Yes"<<endl;
    rep(i,1,n+1)c[i]+=c[i-1],d[i]+=d[i-1];
    int x=0;rep(i,1,n+1)x=max(x,c[i]-d[i-1]);
    REP(i,n)cout<<(i?" ":"")<<b[(i+n-x)%n];cout<<endl;
}

int main()
{
    cin.tie(0);
	ios::sync_with_stdio(false);
    solve();
    return 0;
}