Let us call two integers x and y adjacent if (lcm(x, y))/(gcd(x, y)) is a perfect square. For example, 3 and 12 are adjacent, but 6 and 9 are not.

Here gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y, and lcm(x, y) denotes the [least common multiple (LCM)](https://en.wikipedia.org/wiki/Least_common_multiple) of integers x and y.

You are given an array a of length n. Each second the following happens: each element a_i of the array is replaced by the product of all elements of the array (including itself), that are adjacent to the current value. 

Let d_i be the number of adjacent elements to a_i (including a_i itself). The beauty of the array is defined as max_{1 ≤ i ≤ n} d_i. 

You are given q queries: each query is described by an integer w, and you have to output the beauty of the array after w seconds.

Input

The first input line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.

The following line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^6) — array elements. 

The next line contain a single integer q (1 ≤ q ≤ 3 ⋅ 10^5) — the number of queries.

The following q lines contain a single integer w each (0 ≤ w ≤ 10^{18}) — the queries themselves.

It is guaranteed that the sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values q over all test cases does not exceed 3 ⋅ 10^5

Output

For each query output a single integer — the beauty of the array at the corresponding moment.

Example

Input


2
4
6 8 4 2
1
0
6
12 3 20 5 80 1
1
1


Output


2
3

Note

In the first test case, the initial array contains elements [6, 8, 4, 2]. Element a_4=2 in this array is adjacent to a_4=2 (since (lcm(2, 2))/(gcd(2, 2))=2/2=1=1^2) and a_2=8 (since (lcm(8,2))/(gcd(8, 2))=8/2=4=2^2). Hence, d_4=2, and this is the maximal possible value d_i in this array.

In the second test case, the initial array contains elements [12, 3, 20, 5, 80, 1]. The elements adjacent to 12 are \{12, 3\}, the elements adjacent to 3 are \{12, 3\}, the elements adjacent to 20 are \{20, 5, 80\}, the elements adjacent to 5 are \{20, 5, 80\}, the elements adjacent to 80 are \{20, 5, 80\}, the elements adjacent to 1 are \{1\}. After one second, the array is transformed into [36, 36, 8000, 8000, 8000, 1].
#include <bits/stdc++.h>

using namespace std;

// mt19937 g(chrono::high_resolution_clock::now().time_since_epoch().count());

typedef long long ll;

const int NMAX = 1e6 + 5;
int who[NMAX], fq[NMAX], done[NMAX];

void pre() {
    int N = 1e6;
    for(int i = 1; i <= N; i++) who[i] = i;
    for(int i = 2; i <= N; i++)
        if(who[i] == i && N / i >= i) {
            int val = i * i;
            for(int j = val; j <= N; j += val)
                while(who[j] % val == 0)
                    who[j] /= val;
        }
}

void solve_test() {
    int N;
    scanf("%d", &N);
    vector<int> v(N);
    for(int i = 1; i <= N; i++) {
        int x;
        scanf("%d", &x);
        v[i - 1] = x;
        fq[ who[x] ]++;
    }

    int ans0 = 0;
    int ans1 = fq[1]; done[1] = 1;
    for(int i = 0; i < N; i++) {
        int x = who[ v[i] ];
        ans0 = max(ans0, fq[x]);
        if(!done[x]) {
            done[x] = 1;
            if(fq[x] % 2 == 0)  ans1 += fq[x];
        }
    }

    ans1 = max(ans1, ans0);
    int Q;
    scanf("%d", &Q);
    for(int i = 0; i < Q; i++) {

        ll x;
        scanf("%lld", &x);
        if(x == 0)
            printf("%d\n", ans0);
        else
            printf("%d\n", ans1);
    }

    for(int i = 0; i < N; i++) {
        int x = who[ v[i] ];
        fq[x] = done[x] = 0;
    }
}

int main() {
    //freopen("1.in", "r", stdin);
    //freopen("1.out", "w", stdout);

    pre();

    int T = 1;
    scanf("%d", &T);
    for(int t = 1; t <= T; t++) {
        //printf("Case #%d: ", t);
        solve_test();
    }

    return 0;
}