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 PyPy3 (7.3.0) (judge verdict: Accepted)

N = int(input())
A = [int(x) for x in input().split()]
B = [int(x) for x in input().split()]

cntA = [0 for i in range(N+1)]
cntB = [0 for i in range(N+1)]

for i in range(N):
    cntA[A[i]] += 1
for i in range(N):
    cntB[B[i]] += 1

temp = False
for i in range(N):
    if cntA[A[i]] + cntB[A[i]] > N:
        temp = True
        break

if temp:
    print("No")


else:
    print("Yes")
    cA = 0
    cB = 0
    flag = False
    C = [B[N-i-1] for i in range(N)]
    D = [B[N-i-1] for i in range(N)]

    for i in range(1,N+1):
        if cA + cB + cntA[i] + cntB[i] == N:
            break
        elif cA + cB + cntA[i] + cntB[i] < N:
            cA += cntA[i]
            cB += cntB[i]
        elif cA + cB + cntA[i] >= N:
            if cA >= cntB[i]:
                for j in range(cntB[i]):
                    C[j] = i
                for j in range(N-cB-cntB[i]):
                    C[j+cntB[i]] = D[j]
            else:
                for j in range(cA):
                    C[j] = i
                for j in range(N-cB-cntB[i]):
                    C[j+cA] = D[j]
                for j in range(cB):
                    C[j+N-(cB+cntB[i]-cA)] = D[N-1-j]
                for j in range(cntB[i]-cA):
                    C[N-j-1] = i
            break
        else:
            if cntA[i] + cntB[i] <= N-cB:
                for j in range(cntB[i]-(N-cB-(cA+cntA[i]))):
                    C[j] = i
                for j in range(N-(cB+cntB[i])):
                    C[j+cntB[i]-(N-cB-(cA+cntA[i]))] = D[j]
                for j in range(N-cB-(cA+cntA[i])):
                    C[j+cA+cntA[i]] = i
            else:
                for j in range(cA):
                    C[j] = i
                for j in range(N-(cB+cntB[i])):
                    C[j+cA] = D[j]
                for j in range(cB):
                    C[j+N+cA-(cB+cntB[i])] = D[N-j-1]
                for j in range(cntB[i]-cA):
                    C[N-(cntB[i]-cA)+j] = i
            break
    F = [str(x) for x in C]
    print(" ".join(F))