Problem: Soroush and Keshi each have a labeled and rooted tree on n vertices. Both of their trees are rooted from vertex 1.

Soroush and Keshi used to be at war. After endless decades of fighting, they finally became allies to prepare a Codeforces round. To celebrate this fortunate event, they decided to make a memorial graph on n vertices.

They add an edge between vertices u and v in the memorial graph if both of the following conditions hold: 

  * One of u or v is the ancestor of the other in Soroush's tree. 
  * Neither of u or v is the ancestor of the other in Keshi's tree. 



Here vertex u is considered ancestor of vertex v, if u lies on the path from 1 (the root) to the v.

Popping out of nowhere, Mashtali tried to find the maximum clique in the memorial graph for no reason. He failed because the graph was too big. 

Help Mashtali by finding the size of the maximum clique in the memorial graph.

As a reminder, clique is a subset of vertices of the graph, each two of which are connected by an edge.

Input

The first line contains an integer t (1≤ t≤ 3 ⋅ 10^5) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (2≤ n≤ 3 ⋅ 10^5).

The second line of each test case contains n-1 integers a_2, …, a_n (1 ≤ a_i < i), a_i being the parent of the vertex i in Soroush's tree.

The third line of each test case contains n-1 integers b_2, …, b_n (1 ≤ b_i < i), b_i being the parent of the vertex i in Keshi's tree.

It is guaranteed that the given graphs are trees.

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case print a single integer — the size of the maximum clique in the memorial graph.

Example

Input


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


Output


1
4
1
3

Note

In the first and third test cases, you can pick any vertex.

In the second test case, one of the maximum cliques is \{2, 3, 4, 5\}.

In the fourth test case, one of the maximum cliques is \{3, 4, 6\}.

Solution (in PYTHON3):

Solution:
import sys
input = sys.stdin.buffer.readline

class bit:
    def __init__(self, n):
        self.n = n
        self.a = [0]*(n+1)

    def add(self, idx, val):
        idx += 1
        while idx < self.n:
            self.a[idx] += val
            idx += idx & -idx

    def sumr(self, idx):
        idx += 1
        tot = 0
        while idx:
            tot += self.a[idx]
            idx -= idx & -idx

        return tot

    def sum(self, l, r):
        return self.sumr(r) - self.sumr(l-1)

def add(i):
    bit2.add(tin[i], 1)
    bit1.add(tin[i], i)
    bit1.add(tout[i] + 1, -i)

def remove(i):
    bit2.add(tin[i], -1)
    bit1.add(tin[i], -i)
    bit1.add(tout[i] + 1, i)

for _ in range(int(input())):
    n = int(input())

    adj = [[[] for i in range(n+1)] for j in range(2)]

    for k in range(2):
        a = list(map(int,input().split()))

        for i in range(2,n+1):
            adj[k][a[i-2]].append(i)

    ans = 0

    # we use tin and tout so we can determine if a node is an ancestor of another
    tin = [0]*(n+1)
    tout = [0]*(n+1)
    t = -1
    s = [1]
    visited = [0]*(n+1)

    while s:
        c = s[-1]

        if not visited[c]:
            t += 1
            tin[c] = t

            visited[c] = 1

            for ne in adj[1][c]:
                if not visited[ne]:
                    s.append(ne)
        else:
            tout[c] = t

            s.pop()

    curr_mx = 0
    # we use the bits to keep track of the maximum set of nodes such that no node
    # is an ancestor of another in keshi's tree
    # the tin and tout form disjoint ranges and we keep the max width range if they intersect
    bit1 = bit(n+1)
    bit2 = bit(n+1)
    ops = []  # to undo operations to bits [u,v] -> u was removed v was added

    s = [1]
    visited = [0] * (n + 1)

    while s:
        c = s[-1]

        if not visited[c]:
            visited[c] = 1

            # update set of max elements (implicitly represented by bit1 and bit2)
            u = bit1.sumr(tin[c])
            if u:
                # c is in the subtree of one of the nodes in the set, which we call u.
                # range for c is smaller so add c and remove u
                remove(u)
                add(c)
                ops.append([u,c])

            elif bit2.sum(tin[c],tout[c]):
                # c is the ancestor of one of the nodes in the set
                # range for c is larger so don't add it
                ops.append([0, 0])
            else:
                # c is a new disjoint range so add it
                add(c)
                curr_mx += 1
                ops.append([0,c])

            ans = max(ans, curr_mx)

            for ne in adj[0][c]:
                if not visited[ne]:
                    s.append(ne)
        else:
            u,v = ops.pop()

            if u:
                add(u)
                curr_mx += 1
            if v:
                remove(v)
                curr_mx -= 1

            s.pop()

    print(ans)