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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
    // 一番多いもの、reverseした後一番多いものをずらす、あとはずらすのは一箇所のみ
    int N;
    cin >> N;
    vector<int> A(N);
    rep(i, N) cin >> A[i];
    vector<int> B(N);
    rep(i, N) cin >> B[i];
    vector<int> Acnt(N+1);
    vector<int> Bcnt(N+1);
    rep(i, N) {
        if (Acnt[A[i]] == 0) {
            int x = upper_bound(A.begin(), A.end(), A[i]) - lower_bound(A.begin(), A.end(), A[i]);
            Acnt[A[i]] = x;
        }
        if (Bcnt[B[i]] == 0) {
            int y = upper_bound(B.begin(), B.end(), B[i]) - lower_bound(B.begin(), B.end(), B[i]);
            Bcnt[B[i]] = y;
        }
    }
    int ma = 0;
    int max_cnt = 0;
    bool flag = false;
    for (int i = 1; i <= N; i++) {
        if (Acnt[i] > 0 && Bcnt[i] > 0) flag = true;
        int x = Acnt[i] + Bcnt[i];
        if (x > N) {
            puts("No");
            return 0;
        }
        if (x > max_cnt) {
            max_cnt = x;
            ma = i;
        }
    }
    if (!flag) {
        puts("Yes");
        show(B);
        return 0;
    }
    int A_left = lower_bound(A.begin(), A.end(), ma) - A.begin();
    int A_right = upper_bound(A.begin(), A.end(), ma) - A.begin() - 1;
    reverse(B.begin(), B.end());
    int B_right = (N-1) - (lower_bound(B.rbegin(), B.rend(), ma) - B.rbegin());
    int B_left = (N-1) - (upper_bound(B.rbegin(), B.rend(), ma) - B.rbegin() - 1);
    vector<int> ans(N);
    int cur = A_right + 1;
    int b = B_left;
    rep(i, N) {
        ans[cur] = B[b];
        cur = (cur+1)%N;
        b = (b+1)%N;
    }
    rep(i, N) {
        if (A[i] == ans[i]) {
            while (ans[i] == ans[A_left]) A_left = (A_left+1) % N;
            swap(ans[i], ans[A_left]);
            A_left++;
        }
    }
    puts("Yes");
    show(ans);
    return 0;
    // ll, 0, -, 1i, for s&g, debug
}