Problem: A Pythagorean triple is a triple of integer numbers (a, b, c) such that it is possible to form a right triangle with the lengths of the first cathetus, the second cathetus and the hypotenuse equal to a, b and c, respectively. An example of the Pythagorean triple is (3, 4, 5).

Vasya studies the properties of right triangles, and he uses a formula that determines if some triple of integers is Pythagorean. Unfortunately, he has forgotten the exact formula; he remembers only that the formula was some equation with squares. So, he came up with the following formula: c = a^2 - b.

Obviously, this is not the right formula to check if a triple of numbers is Pythagorean. But, to Vasya's surprise, it actually worked on the triple (3, 4, 5): 5 = 3^2 - 4, so, according to Vasya's formula, it is a Pythagorean triple.

When Vasya found the right formula (and understood that his formula is wrong), he wondered: how many are there triples of integers (a, b, c) with 1 ≤ a ≤ b ≤ c ≤ n such that they are Pythagorean both according to his formula and the real definition? He asked you to count these triples.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Each test case consists of one line containing one integer n (1 ≤ n ≤ 10^9).

Output

For each test case, print one integer — the number of triples of integers (a, b, c) with 1 ≤ a ≤ b ≤ c ≤ n such that they are Pythagorean according both to the real definition and to the formula Vasya came up with.

Example

Input


3
3
6
9


Output


0
1
1

Note

The only Pythagorean triple satisfying c = a^2 - b with 1 ≤ a ≤ b ≤ c ≤ 9 is (3, 4, 5); that's why the answer for n = 3 is 0, and the answer for n = 6 (and for n = 9) is 1.

Solution (in CPP):

Solution:
// Problem: D. Pythagorean Triples
// Contest: Codeforces - Educational Codeforces Round 104 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1487/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ld long double
#define pll pair<ll, ll>
#define X first
#define Y second
#define pb emplace_back
#define nl "\n"
#define vl vector<ll>
#define mod 1000000007

ll dx[] = {0, 1, 0, -1, 0};

ll min(ll a, ll b) { return a < b ? a : b; }
ll max(ll a, ll b) { return a > b ? a : b; }
ll power(ll a, ll b, ll m) {
  a %= m;
  ll r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return r;
}

vl dp;

void solve() {
  // code...
  ll n;
  cin >> n;
  ll i = sqrt(2 * n - 1);
  cout << dp[i - 1] - 1 << nl;
  return;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll i = 1;
  for (; (i * i + 1) / 2 <= 1e9; i++) {
    if ((i * i) & 1) {
      dp.pb(1);
    } else {
      dp.pb(0);
    }
  }
  for (ll i = 1; i < dp.size(); i++) {
    dp[i] += dp[i - 1];
  }
  ll t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}