Given n integers you cangenerate 2ⁿ-1 non-empty subsets from them. Determine for howmany of these subsets the product of all the integers in that is a perfectsquare. For example for the set {4,6,10,15} there are3
such subsets. {4}, {6,10,15} and {4,6,10,15}. Aperfect square is an integer whose square root is an integer. For example 1, 4,9, 16,…. .

Input

Input contains multiple testcases. First line of the input contains T(1≤T≤30)the number of test cases. Each test case consists of 2 lines. First line containsn(1≤n≤100) and second linecontains
n space separated integers. All these integers are between 1 and 10^15.  None of these integers is divisible by aprime greater than 500.

Output

For each test caseoutput is a single line containing one integer denoting the number of non-emptysubsets whose integer product is a perfect square. The input will be such thatthe result will always fit into signed 64 bit integer.

SampleInput                              Output for Sample Input

Problemsetter: Abdullah al Mahmud

SpecialThanks to: Manzurur RahmanKhan

题意：给出n个正整数，从中选出1个或者多个，使得选出来的整数乘积是完全平方数，一共有多少种选法。

思路：用01向量表示一个数，再用n个01向量来表示我们的选择，因为完全平方数要求素因子的次数一定要是偶数的，所以我们可以统计的将奇数当作1，偶数当作0，那么就是一组可以变换成oxr的方程组，最后的结果有自由变量f个，答案是2^f-1,f求解就是求n-方程组的秩

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
typedef long long ll;
using namespace std;
const int maxn = 510;

typedef int Matrix[maxn][maxn];
int prime[maxn], vis[maxn];
Matrix A;

int gen_primes(int m) {
	memset(vis, 0, sizeof(vis));
	int cnt = 0;
	for (int i = 2; i < m; i++) {
		if (!vis[i]) {
			prime[cnt++] = i;
			for (int j = i * i; j < m; j += i)
				vis[j] = 1;
		}
	}
	return cnt;
}

int rank(Matrix A, int m, int n) {
	int i = 0, j = 0, k , r, u;
	while (i < m && j < n) {
		r = i;
		for (k = i; k < m; k++)
			if (A[k][j]) {
				r = k;
				break;
			}
		if (A[r][j]) {
			if (r != i)
				for (k = 0; k <= n; k++)
					swap(A[r][k], A[i][k]);
			for (u = i+1; u < m; u++)
				if (A[u][j])
					for (k = i; k <= n; k++)
						A[u][k] ^= A[i][k];
			i++;
		}
		j++;
	}
	return i;
}

int main() {
	int m = gen_primes(505);

	int t;
	scanf("%d", &t);
	while (t--) {
		int n, maxp = 0;;
		ll x;
		scanf("%d", &n);

		memset(A, 0, sizeof(A));
		for (int i = 0; i < n; i++) {
			scanf("%lld", &x);
			for (int j = 0; j < m; j++)
				while (x % prime[j] == 0) {
					maxp = max(maxp, j);
					x /= prime[j];
					A[j][i] ^= 1;
				}
		}

		int r = rank(A, maxp+1, n);
		printf("%lld\n", (1ll << (n-r)) - 1);
	}
	return 0;
}