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. 
//package codeforces.edu111;

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

import static java.lang.Math.*;

//Think through the entire logic before jump into coding!
//If you are out of ideas, take a guess! It is better than doing nothing!
//Read both C and D, it is possible that D is easier than C for you!

//Be aware of integer overflow!
//If you find an answer and want to return immediately, don't forget to flush before return!

public class A {
    static InputReader in;
    static PrintWriter out;

    public static void main(String[] args) {
        //initReaderPrinter(true);
        initReaderPrinter(false);
        solve(in.nextInt());
        //solve(1);        
    }

    static void solve(int testCnt) {
        for (int testNumber = 0; testNumber < testCnt; testNumber++) {
            int s = in.nextInt();
            int k = 1;
            while(k * k < s) {
                k++;
            }
            out.println(k);
        }
        out.close();
    }

    static void initReaderPrinter(boolean test) {
        if (test) {
            try {
                in = new InputReader(new FileInputStream("src/input.in"));
                out = new PrintWriter(new FileOutputStream("src/output.out"));
            } catch (IOException e) {
                e.printStackTrace();
            }
        } else {
            in = new InputReader(System.in);
            out = new PrintWriter(System.out);
        }
    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        InputReader(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream), 32768);
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                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());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        Integer[] nextIntArray(int n) {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArrayPrimitive(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArrayPrimitiveOneIndexed(int n) {
            int[] a = new int[n + 1];
            for (int i = 1; i <= n; i++) a[i] = nextInt();
            return a;
        }

        Long[] nextLongArray(int n) {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long[] nextLongArrayPrimitive(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long[] nextLongArrayPrimitiveOneIndexed(int n) {
            long[] a = new long[n + 1];
            for (int i = 1; i <= n; i++) a[i] = nextLong();
            return a;
        }

        String[] nextStringArray(int n) {
            String[] g = new String[n];
            for (int i = 0; i < n; i++) g[i] = next();
            return g;
        }
    }
}