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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)
#define VL vector<ll>
#define VS vector<string>
#define VB vector<bool>
#define VP vector<pair<ll,ll>>
#define VVL vector<vector<ll>>
#define VVP vector<vector<pair<ll,ll>>>
#define PL pair<ll,ll>
#define ALL(v) (v).begin(), (v).end()
ll d1[4] = {1, -1, 0, 0};
ll d2[4] = {0, 0, 1, -1};

int main() {
    ll n;
    cin >> n;
    VL a(n), b(n);
    vector<ll> ma(n+1, 0);
    rep(i, 0, n) cin >> a[i];
    rep(i, 0, n) cin >> b[i];
    rep(i, 0, n){
        ma[a[i]]++;
    }
    VP cp(n);
    rep(i, 0, n) {
        cp[i].first = a[i];
        cp[i].second = i;
    }
    sort(a.begin(), a.end(), [&](auto const& l, auto const& r){
        if(ma[l] > ma[r]) return true;
        else if(l < r) return true;
        else return false;
    });
    sort(cp.begin(), cp.end(), [&](auto const& l, auto const& r){
        if(ma[l.first] > ma[r.first]) return true;
        else if(l.first < r.first) return true;
        else return false;
    });
    sort(b.begin(), b.end(), [&](auto const& l, auto const& r){
        if(ma[l] > ma[r]) return true;
        else if(l < r) return true;
        else return false;
    });
    
    //rep(i, 0, n) cerr << cp[i].first << (i == n-1 ? "\n" : " ");
    //rep(i, 0, n) cerr << a[i] << (i == n-1 ? "\n" : " ");
    //cerr << "a" << endl;

    VL ans(n);

    ll srt = -1;
    rep(i, 0, n) {
        if(a[0] != b[i]) {
            srt = i;
            break;
        }
    }
    if(srt == -1) {
        cout << "No" << endl;
        return 0;
    }
    rep(i, 0, n) ans[i] =  b[(i + srt) % n];
    rep(i, 0, n) {
        if(a[i] == ans[i]) {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;
    //rep(i, 0, n) cerr << ans[i] << (i == n-1 ? "\n" : " ");
    VL true_ans(n);
    rep(i, 0, n) true_ans[cp[i].second] = ans[i];
    rep(i, 0, n) cout << true_ans[i] << (i == n-1 ? "\n" : " ");
    return 0;
}