You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.

Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).

We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k  times}.

Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).

It can be proved that the answer always exists.

Input

The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.

Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.

The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.

The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.

It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.

Output

Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.

Example

Input


3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4


Output


1
5
2

Note

In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.

In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.

In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from math import sqrt

t=int(input())
for tests in range(t):
    n=int(input())
    P=list(map(int,input().split()))
    C=list(map(int,input().split()))

    for i in range(n):
        P[i]-=1

    USE=[0]*n
    ANS=n

    for i in range(n):
        if USE[i]==1:
            continue

        cy=i
        CYCLE=[]

        while USE[cy]!=1:
            USE[cy]=1
            cy=P[cy]
            CYCLE.append(cy)

        #print(CYCLE)

        LEN=len(CYCLE)
        x=LEN
        xr=int(sqrt(x))+1
        LIST=[]

        for k in range(1,xr+1):
            if x%k==0:
                LIST.append(k)
                LIST.append(x//k)

        for l in LIST:
            if ANS<=LEN//l:
                continue
            COLORS=[-1]*(LEN//l)
            ind=0
            for k in range(LEN):
                if COLORS[ind]==-1:
                    COLORS[ind]=C[cy]
                elif COLORS[ind]==C[cy]:
                    True
                else:
                    COLORS[ind]=0
                    
                cy=P[cy]
                ind+=1
                if ind==LEN//l:
                    ind=0

            for c in COLORS:
                if c!=0:
                    ANS=LEN//l
                    break

            #print(CYCLE,l,COLORS)

    print(ANS)

            

        
            
        
