模拟前几个样例可以发现需要根据奇偶性讨论。反向考虑，当结果为 $0, 1, 2, 3, 4$ 的时候的节点分配是怎样的，可以得到当答案 $k$ 是奇数的时候，能够放下的最多的点数是 $2(k + 1)(\frac{k}{2} + 1)$；当答案 $k$ 是偶数的时候，能够放下的最多的点数是 $(k + 1)^2$。然后就可以二分答案了。

// Date: Thu Dec  7 21:25:48 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

int t;
ll n;

bool check(ll x) {
  ll cnt;

  if (x & 1) {
    cnt = (x + 1) * (x / 2 + 1) * 2;
  } else {
    cnt = (1 + x) * (1 + x);
  }

  return cnt < n;
}

int main(void) {
#ifdef _DEBUG
  freopen("1809b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    if (n == 1) {
      cout << "0\n";
      continue;
    }

    ll l = 0, r = 1e10, mid;
    while (l < r) {
      mid = (l + r + 1) / 2;
      if (check(mid))
        l = mid;
      else
        r = mid - 1;
    }

    cout << l + 1 << '\n';
  }

  return 0;
}