Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

Yes
2 2 3 1 1 1

Sample Input 2

3
1 1 2
1 1 3

Sample Output 2

No

Sample Input 3

4
1 1 2 3
1 2 3 3

Sample Output 3

Yes
3 3 1 2

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define pi acos(-1.0)
#define pb push_back
#define mp make_pair
#define ll long long
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define testcase ll T; cin >> T; for (ll tc = 1; tc <= T; tc++)
#define M 1000000007
#define MM 2000000000000000000
#define eps 1e-8
#define eq(x,y) (fabs((x)-(y)) < eps)
#define r2 1.41421356237
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod;assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

int main()
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  //testcase {
  int n,p;
  cin>>n;
  bool ok=0;
  vector<int>v1(n),v2(n),v3(n);
  for(int i=0;i<n;i++) { cin>>v1[i]; }
  for(int i=0;i<n;i++) { cin>>v2[i]; }
  for(int k=0;k<=1000;k++) {
  	ok=0;
  	for(int i=0;i<n;i++) {
  		if(v1[i]!=v2[(i+k)%n]) { ok=1; }
  		else { ok=0; break; }
  	}
  	if(ok) { p=k; break; }
  }
  if(!ok) { cout<<"No"; }
  else {
  	cout<<"Yes\n";
  	for(int i=0;i<n;i++) { v3[i]=v2[(i+p)%n]; }
for(auto x:v3) { cout<<x<<' '; }
}
  //}
  return 0;
}