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.util.*;
import java.io.*;

public class EC111A {
   public static void main(String[] args) {
      FastIO io = new FastIO();
      int array[] = new int[80];
      int temp = 1;
      int current = 1;
      array[0] = 0;
      array[1] = temp;
      int count = 2; 
      while (temp <= 5500) {
         current += 2;
         temp = temp + current;
         array[count] = temp;
         count++;
      }
      
      int t = io.nextInt();
      
      while (t --> 0) {
         int s = io.nextInt();
         for (int i=0; i<array.length; i++) {  
            if (s > array[i] && s <= array[i+1]) {
               io.println(i+1);
               break;
            }
         }
      }
      io.close();
   }
   public static int[] BS1(int array[], int left, int right, int target) {
      //Search where a value belongs between 2 values
      //Also can be used for > or <, change return type to 1 value
      int[] answer = new int[2];
      if (target < array[0]) {
         answer[0] = -1;
         answer[1] = 0;
         return answer;
      }
      if (target > array[array.length-1]) {
         answer[0] = array.length-1;
         answer[1] = array.length;
         return answer;
      }
      
      
      while (left <= right) {
         int mid = (left + right >> 1) + (left + right & 1);
         if ((array[mid] == target) || (array[mid] < target && array[mid+1] > target)) {
            answer[0] = mid;
            answer[1] = mid+1;
            return answer;
         }
         else if (array[mid] < target) {
            left = mid;
         }
         else if (array[mid] > target) {
            right = mid-1;
         }
      }
      return answer;
   }
   static class FastIO extends PrintWriter {
        BufferedReader br;
        StringTokenizer st;
 
        public FastIO() {
            super(System.out);
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        public String next() {
            try {
                while (st == null || !st.hasMoreTokens()) {
                    st = new StringTokenizer(br.readLine());
                }
                return st.nextToken();
            } catch (Exception e) {
                return null;
            }
        }
 
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
        public double nextDouble() {
            return Double.parseDouble(next());
        }
 
        public long nextLong() {
            return Long.parseLong(next());
        }
 
        public void close() {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
            super.close();
        }
    }
}