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>
#include <vector>
#include <set>
#include <string>
#include <map>
#include <cmath>
#include <algorithm>
typedef long long int ll;
#define f first
#define s second
#define vi vector<int>
#define ld long double
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define floop(count,a, n) for(int count{a}; count < n; ++count)
#define tr(c, it) for(auto it{c.begin()}; it != c.end(); ++it)
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define newl '\n'



using namespace std;

const int MOD = 1e9+7;
const ld PI = acos(-1);
const ld EPS = 1e-9;
const ll INF = 1e18;

    
int main(){

    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
        #ifndef ONLINE_JUDGE       
        freopen("input.txt", "r", stdin);                                     
        freopen("output.txt", "w", stdout);                       
        #endif    
    int cas;cin>>cas;
        while(cas--){
            int n;
            cin>>n;
            int a = sqrt(n);
            if(a*a == n)
                cout<<a<<"\n";
            else
                cout<<a + 1<<"\n";


        }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";

    return 0; 
}