Codeforces 1744D. Divisibility by 2^n

D. Divisibility by 2^n
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an array of positive integers a1,a2,…,an.

Make the product of all the numbers in the array (that is, a1⋅a2⋅…⋅an) divisible by 2n.

You can perform the following operation as many times as you like:

select an arbitrary index i (1≤i≤n) and replace the value ai with ai=ai⋅i.
You cannot apply the operation repeatedly to a single index. In other words, all selected values of i must be different.

Find the smallest number of operations you need to perform to make the product of all the elements in the array divisible by 2n. Note that such a set of operations does not always exist.

Input
The first line of the input contains a single integer t (1≤t≤104) — the number test cases.

Then the descriptions of the input data sets follow.

The first line of each test case contains a single integer n (1≤n≤2⋅105) — the length of array a.

The second line of each test case contains exactly n integers: a1,a2,…,an (1≤ai≤109).

It is guaranteed that the sum of n values over all test cases in a test does not exceed 2⋅105.

Output
For each test case, print the least number of operations to make the product of all numbers in the array divisible by 2n. If the answer does not exist, print -1.

Example
inputCopy
6
1
2
2
3 2
3
10 6 11
4
13 17 1 1
5
1 1 12 1 1
6
20 7 14 18 3 5
outputCopy
0
1
1
-1
2
1
Note
In the first test case, the product of all elements is initially 2, so no operations needed.

In the second test case, the product of elements initially equals 6. We can apply the operation for i=2, and then a2 becomes 2⋅2=4, and the product of numbers becomes 3⋅4=12, and this product of numbers is divided by 2n=22=4.

In the fourth test case, even if we apply all possible operations, we still cannot make the product of numbers divisible by 2n — it will be (13⋅1)⋅(17⋅2)⋅(1⋅3)⋅(1⋅4)=5304, which does not divide by 2n=24=16.

In the fifth test case, we can apply operations for i=2 and i=4.

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5;
int main(void) {
	int t;
	cin >> t;
	int n, x;
	while (t--) {
		cin >> n;
		vector<int>v(n);
		int cnt = 0, cnt1;
		for (int i = 1; i <= n; i++) {
			cin >> x;
			cnt1 = 0;
			while (x % 2 == 0) {
				cnt++;
				x /= 2;
			}
			int j = i;
			while (j % 2 == 0) {
				cnt1++;
				j /= 2;
			}
			if (cnt1 != 0)v.push_back(cnt1);
		}
		sort(v.begin(), v.end());
		int ans = 0;
		for (int i = v.size() - 1; i >= 0 && cnt < n; i--) {
			cnt += v[i];
			ans++;
		}
		if (cnt < n)cout << -1 << endl;
		else cout << ans << endl;
	}
}
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