Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.

For example: 

  * the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds; 
  * the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array; 
  * the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1; 
  * the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1; 
  * the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array. 



You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.

Input

The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.

Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.

Output

Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.

Example

Input


4
1
8
7
42


Output


1
3
3
7

Note

Consider the example test:

  1. in the first test case, the array [1] meets all conditions; 
  2. in the second test case, the array [3, 4, 1] meets all conditions; 
  3. in the third test case, the array [1, 2, 4] meets all conditions; 
  4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. 
#include <bits/stdc++.h>
using namespace std;
#define FAST1 ios_base::sync_with_stdio(false)
#define FAST2 cin.tie(NULL)
#define ll long long int
#define vctl vector<long long int>
#define vlrl valarray<long long int>
#define f(i, init, n) for (ll i = init; i < n; i++)
#define rf(i, n, end) for (ll i = n; i > end; i--)
#define all(a) a.begin(), a.end()
#define mii map<long long int, long long int>
#define pii pair<long long int, long long int>
#define mod 1000000007
ll power(ll x, ll y)
{
    if (y == 0)
        return 1;
    else if (y == 1)
        return x;
    else if (y % 2 == 0)
        return power(x, y / 2) * power(x, y / 2);
    else
        return x * power(x, y / 2) * power(x, y / 2);
}
void printVector(vctl &v)
{
    for (auto j : v)
        cout << j << " ";
    cout << endl;
}
int main()
{
    FAST1;
    FAST2;
    ll tc = 1;
    cin >> tc;
    while (tc--)
    {
        int s;
        cin >> s;
        int curr = 0;
        int i = 1;
        int ans = 0;
        while (curr < s)
        {
            curr += i;
            i += 2;
            ans++;
        }
        cout << ans << endl;
    }
    return 0;
}