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: Runtime Error)

import kotlin.math.*

fun main() {
    println(solve())
}

val solve: () -> Any = solve@{
    val N = readInt()
    val A = readIntList()
    val (nums, counts) = readIntList()
        .groupingBy { it }
        .eachCount()
        .toList()
        .sortedBy { it.second }
        .unzip()
        .let { it.first.toMutableList() to it.second.toMutableList() }
    val res = mutableListOf<Int>()
    for (a in A) {
        val f = nums.first()
        if (a != f) {
            res.add(f)
            counts[0]--
            if (counts[0] == 0)  {
                nums.removeAt(0)
                counts.removeAt(0)
            }
        }
        else {
            if (nums.isEmpty()) return@solve "No"
            res.add(counts[1])
            counts[1]--
            if (counts[1] == 0){
                nums.removeAt(1)
                counts.removeAt(1)
            }
        }
    }
    "Yes\n"+res.toList().joinToString { " " }+"\n"
//    res.joinToString { " " }
}

val mod: Int = 1_000_000_007


fun readString() = readLine()!!
fun readStringList() = readLine()!!.split(" ")
fun readInt() = readLine()!!.toInt()
fun readLong() = readLine()!!.toLong()
fun readIntList() = readLine()!!.split(" ").map(String::toInt)
fun readLongList() = readLine()!!.split(" ").map(String::toLong)

fun <T> T.alsoPrintln() = also { println(this) }

fun Boolean.toYesOrNo(yes: String = "Yes", no: String = "No") = if (this) yes else no

fun <E> List<E>.toPair() = this[0] to this[1]

fun <T> readPair(t: (String) -> T) = readPair(t, t)
fun <S, T> readPair(s: (String) -> S, t: (String) -> T) =
    readLine()!!.split(" ").let { Pair(s(it[0]), t(it[1])) }

fun Int.pow(n: Int) = this.toDouble().pow(n).toInt()

fun minAndMax(a: Int, b: Int) = if (a < b) a to b else b to a

fun <T> T.ifOrThis(bool: (T) -> Boolean, trans: (T) -> T) = if (bool(this)) trans(this) else this
fun <T> T.ifOrThis(bool: Boolean, trans: (T) -> T) = if (bool) trans(this) else this