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. 
#include<bits/stdc++.h>
#define max3(a,b,c) max(max((a),(b)),(c))
#define max4(a,b,c,d) max(max((a),(b)),max((c),(d)))
#define min3(a,b,c) min(min((a),(b)),(c))
#define min4(a,b,c,d) min(min((a),(b)),min((c),(d)))
#define INF 1e18
#define debug(x) cout << #x << " = " << x << endl;
#define endl "\n"
#define ll long long 
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
using namespace std;

  void solve(){
     int s;
     cin>>s;
     bool k=false;
     int h=1;
     int l=1;
     int ct=1;
      while(h<s){
        if(h==s){
          k=true;
          break;
        }else if(h+l+1==s || h+l+2==s){
          ct++;
          break;
        }
        l=l+2;
        h=h+l;
         ct++;
      }
      cout<<ct<<endl;
 }
 
 int main(){
    ios_base::sync_with_stdio(false);           
    cin.tie(NULL);
     ll t=1;
     cin>>t;
     while(t--){
      solve();
     }
      return 0;
}