#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 3;
const int INF = 0x3f3f3f3f;
bool flag[N];
long long ans[N] = {0, 1};
vector<int> vec;
long long gcd(long long p, long long q) { return q ? gcd(q, p % q) : p; }
bool triangle(long long a, long long b, long long c) {
  return a + b > c && b + c > a && c + a > b;
}
bool isPrime(long long N) {
  if (N <= 1) return false;
  for (int i = 2; i <= sqrt(N); ++i)
    if (N % i == 0) return false;
  return true;
}
int main() {
  for (int i = 2; i < (int)1e6; ++i) {
    if (isPrime(i)) vec.push_back(i);
  }
  int preMax = 1, preVecP = 0, preP = 0;
  long long t, n;
  cin >> t;
  while (t--) {
    scanf("%lld", &n);
    if (n <= preMax)
      printf("%lld\n", ans[n]);
    else {
      int x;
      for (int i = preMax + 1; i <= n; ++i) {
        if (i == vec[preVecP]) {
          ans[i] = ans[i - 1] + 1;
          preVecP++;
        } else if (i % vec[preP] == 0 && (x = gcd(i, vec[preP])) &&
                   triangle(x, i / x, vec[preP] / x)) {
          ans[i] = ans[i - 1] - 1;
          preP++;
        } else {
          ans[i] = ans[i - 1];
        }
      }
      printf("%lld\n", ans[n]);
      preMax = n;
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

In number world, two different numbers are friends if they have a lot in common, but also each one has unique perks.

More precisely, two different numbers a and b are friends if gcd(a,b), (a)/(gcd(a,b)), (b)/(gcd(a,b)) can form sides of a triangle.

Three numbers a, b and c can form sides of a triangle if a + b > c, b + c > a and c + a > b.

In a group of numbers, a number is lonely if it doesn't have any friends in that group.

Given a group of numbers containing all numbers from 1, 2, 3, ..., n, how many numbers in that group are lonely?

Input

The first line contains a single integer t (1 ≤ t ≤ 10^6) - number of test cases.

On next line there are t numbers, n_i (1 ≤ n_i ≤ 10^6) - meaning that in case i you should solve for numbers 1, 2, 3, ..., n_i.

Output

For each test case, print the answer on separate lines: number of lonely numbers in group 1, 2, 3, ..., n_i.

Example

Input


3
1 5 10


Output


1
3
3

Note

For first test case, 1 is the only number and therefore lonely.

For second test case where n=5, numbers 1, 3 and 5 are lonely.

For third test case where n=10, numbers 1, 5 and 7 are lonely.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 3;
const int INF = 0x3f3f3f3f;
bool flag[N];
long long ans[N] = {0, 1};
vector<int> vec;
long long gcd(long long p, long long q) { return q ? gcd(q, p % q) : p; }
bool triangle(long long a, long long b, long long c) {
  return a + b > c && b + c > a && c + a > b;
}
bool isPrime(long long N) {
  if (N <= 1) return false;
  for (int i = 2; i <= sqrt(N); ++i)
    if (N % i == 0) return false;
  return true;
}
int main() {
  for (int i = 2; i < (int)1e6; ++i) {
    if (isPrime(i)) vec.push_back(i);
  }
  int preMax = 1, preVecP = 0, preP = 0;
  long long t, n;
  cin >> t;
  while (t--) {
    scanf("%lld", &n);
    if (n <= preMax)
      printf("%lld\n", ans[n]);
    else {
      int x;
      for (int i = preMax + 1; i <= n; ++i) {
        if (i == vec[preVecP]) {
          ans[i] = ans[i - 1] + 1;
          preVecP++;
        } else if (i % vec[preP] == 0 && (x = gcd(i, vec[preP])) &&
                   triangle(x, i / x, vec[preP] / x)) {
          ans[i] = ans[i - 1] - 1;
          preP++;
        } else {
          ans[i] = ans[i - 1];
        }
      }
      printf("%lld\n", ans[n]);
      preMax = n;
    }
  }
  return 0;
}

```