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. 

import java.io.*;
import java.util.*;
import java.lang.*;
import static java.lang.Math.*;

// Sachin_2961 submission //
public class Codeforces {
    public void solve() {
        int n = fs.nInt();
        int sum = 1,cnt=1,next=sum+2;
        while (sum < n ){
            sum += next;
            next += 2;
            cnt++;
        }
        out.println(cnt);
    }
    static boolean multipleTestCase = true; static FastScanner fs; static PrintWriter out;
    public void run(){
        fs = new FastScanner();
        out = new PrintWriter(System.out);
        int tc = (multipleTestCase)?fs.nInt():1;
        while (tc-->0)solve();
        out.flush();
        out.close();
    }
    public static void main(String[]args){
        try{
            new Codeforces().run();
        }catch (Exception e){
            e.printStackTrace();
        }

    }
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String n() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        String Line()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
        int nInt() {return Integer.parseInt(n()); }
        long nLong() {return Long.parseLong(n());}
        double nDouble(){return Double.parseDouble(n());}
        int[]aI(int n){
            int[]ar = new int[n];
            for(int i=0;i<n;i++)
                ar[i] = nInt();
            return ar;
        }
    }
    public static void sort(int[] arr){
        ArrayList<Integer> ls = new ArrayList<Integer>();
        for(int x: arr)
            ls.add(x);
        Collections.sort(ls);
        for(int i=0; i < arr.length; i++)
            arr[i] = ls.get(i);
    }
}