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 Solution {

    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static boolean isPrime(int n) {
        if (n < 2) return false;
        for (int x = 2; x * x <= n; x++) {
            if (n % x == 0) return false;
        }
        return true;
    }

    static boolean[] sieveOfEratosthenes(int n) {
        boolean[] ans = new boolean[n + 1];
        Arrays.fill(ans, true);
        ans[1] = false;
        for (int i = 2; i <= n; i++) {
            if (ans[i]) {
                for (int j = 2 * i; j <= n; j += i) {
                    ans[j] = false;
                }
            }
        }
        return ans;
    }

    static int modInverse(int a, int m) {
        for (int x = 1; x < m; x++)
            if (((a%m) * (x%m)) % m == 1)
                return x;
        return 1;
    }

    long modPow(int x, int n, int m) {
        if (n == 0) return 1 % m;
        long u = modPow(x, n / 2, m);
        u = (u * u) % m;
        if (n % 2 == 1) u = (u * x) % m;
        return u;
    }

    int max(int ...nums){
        return Arrays.stream(nums).max().getAsInt();
    }

    static class IScanner {
        BufferedReader br;
        StringTokenizer st;

        public IScanner() {
            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());
        }

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

    public static void main(String[] args) {
        try {
            solve();
        }catch (Exception e){
            e.printStackTrace();
        }
    }

    static class Node{
        List<int[]> adj = new ArrayList<>();
    }
    public static void solve(){
        Solution.IScanner in = new IScanner();
        int t = in.nextInt();
        while(t-- > 0){
            int n = in.nextInt();
            int s = 1;
            int len = 0;
            while(n > 0){
                len++;
                n -= s;
                s += 2;
            }

            System.out.println(len);

        }
        PrintWriter out = new PrintWriter(System.out);
        out.flush();

    }
}
