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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import kotlin.math.max

fun main(){
    val n = readLine()!!.split(" ")[0].toInt()
    val a = readLine()!!.split(" ").map { it.toInt() }
    val b = readLine()!!.split(" ").map { it.toInt() }
    var count = 0

    var counttwo = 0

    if (a[0] == b[0]) counttwo++
    for (i in 1 until n){
        if (a[i - 1] != a[i] || b[i - 1] != b[i]) {
            count = max(count, counttwo)
            counttwo = 0

        }
        if (a[i] == b[i]){
            counttwo++
        }
    }
    count = max(count, counttwo)
    val ans = b.sortedDescending().toMutableList()
    var frontflag = 0
    var backflag = n - 1
    for (i in 0 until n){
        if (a[i] == ans[i]){
            val tmp = ans[frontflag]
            ans[frontflag] = ans[i]
            ans[i] = tmp
            if (ans[frontflag] == a[frontflag]){
                val tmpp = ans[frontflag]
                ans[frontflag] = ans[backflag]
                ans[backflag] = tmpp
                backflag--
                frontflag--
            } else if (ans[i] == a[i]){
                val tmpp = ans[i]
                ans[i] = ans[backflag]
                ans[backflag] = tmpp
                backflag--
                frontflag--
            }
            frontflag++
        }
    }
    if (n / 2 >= count){
        println("Yes")
        for (i in 0 until n){
            print("${ans[i]} ")
        }
    } else {
        println("No")
    }
}