Consider some positive integer x. Its prime factorization will be of form x = 2^{k_1} ⋅ 3^{k_2} ⋅ 5^{k_3} ⋅ ...

Let's call x elegant if the greatest common divisor of the sequence k_1, k_2, ... is equal to 1. For example, numbers 5 = 5^1, 12 = 2^2 ⋅ 3, 72 = 2^3 ⋅ 3^2 are elegant and numbers 8 = 2^3 (GCD = 3), 2500 = 2^2 ⋅ 5^4 (GCD = 2) are not.

Count the number of elegant integers from 2 to n.

Each testcase contains several values of n, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (1 ≤ T ≤ 10^5) — the number of values of n in the testcase.

Each of the next T lines contains a single integer n_i (2 ≤ n_i ≤ 10^{18}).

Output

Print T lines — the i-th line should contain the number of elegant numbers from 2 to n_i.

Example

Input

4
4
2
72
10


Output

2
1
61
6

Note

Here is the list of non-elegant numbers up to 10:

  * 4 = 2^2, GCD = 2; 
  * 8 = 2^3, GCD = 3; 
  * 9 = 3^2, GCD = 2. 



The rest have GCD = 1.
// upsolve with rainboy
import java.io.*;
import java.util.*;

public class CF1036F {
	static long[] pp = new long[1037552];
	static int cnt;
	static long sqrt_(long n, long lower) {
		long upper = lower * 2;
		while (upper * upper <= n) {
			lower = upper;
			upper *= 2;
		}
		while (upper - lower > 1) {
			long r = (lower + upper) / 2;
			if (r * r <= n)
				lower = r;
			else
				upper = r;
		}
		return lower;
	}
	static void init() {
		cnt = 0;
		for (int a = 2; a <= 1000000; a++) {
			long p = (long) a * a;
			while (p <= 1000000000000000000L / a) {
				p *= a;
				pp[cnt++] = p;
			}
		}
		Arrays.sort(pp);
		int cnt_ = 0;
		for (int i = 0; i < cnt; i++)
			if (i == 0 || pp[i] != pp[i - 1])
				pp[cnt_++] = pp[i];
		cnt = cnt_;
		cnt_ = 0;
		long r = 1;
		for (int i = 0; i < cnt; i++) {
			r = sqrt_(pp[i], r);
			if (r * r != pp[i])
				pp[cnt_++] = pp[i];
		}
		cnt = cnt_;
	}
	public static void main(String[] args) throws IOException {
		init();
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		int t = Integer.parseInt(br.readLine());
		while (t-- > 0) {
			long n = Long.parseLong(br.readLine());
			int lower = -1, upper = cnt;
			while (upper - lower > 1) {
				int i = (lower + upper) / 2;
				if (pp[i] <= n)
					lower = i;
				else
					upper = i;
			}
			pw.println(n - sqrt_(n, 1) - (lower + 1));
		}
		pw.close();
	}
}
