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.*;
public class Codechef{

    
    public static void main(String args[]){
        FastReader sc=new FastReader();
        int t=sc.nextInt();
        
        
            while(t-->0){
                int n=sc.nextInt();
                ArrayList<Integer>al=new ArrayList<Integer>();
                int sum=0;
                int i=1;
            
                while(sum<n){
                    sum=sum+i;
                    //sum++;
                    i=i+2;

                
                    al.add(1);
                }
                System.out.println(al.size());
        
            }
        }
}

class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader()
    {
        br = new BufferedReader(
            new InputStreamReader(System.in));
    }

    String next()
    {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            }
            catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() { return Integer.parseInt(next()); }

    long nextLong() { return Long.parseLong(next()); }

    double nextDouble()
    {
        return Double.parseDouble(next());
    }
    
    float nextFloat()
    {
        return Float.parseFloat(next());
    }

    String nextLine()
    {
        String str = "";
        try {
            str = br.readLine();
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
    
    int[] readArray(int n) {
        int[] a=new int[n];
        for (int i=0; i<n; i++) a[i]=nextInt();
        return a;
    }
}
    