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. 
                  /**Bismillahir Rahmanir Rahim.**/
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long int   ull;
typedef long long int            ll;
typedef long double              ld;
#define Faster  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pi                 acos(-1)
#define test        int t; cin>>t; while(t--)
#define dot(x)      fixed<<setprecision(x)
#define eps 0.000000001
int chessx[]={-1,-1,1, 1,-2,-2, 2,2}; //knight
int chessy[]={ 2,-2,2,-2, 1,-1,-1,1}; //knight
int xx[]={ 0,0,1,-1};
int yy[]={-1,1,0, 0};
const double EPS = 1e-12;
bool isEqual(ld a, ld b) {
    return abs(a - b) <= EPS;
}
bool isGreater(ld a, ld b) {
    return a > b + EPS;
}
bool isSmaller(ld a, ld b) {
    return a + EPS < b;
}
// cin.ignore();
//sort(s,s+n,greater<int>());
//cin.getline(array,size);
//s.erase(s.begin()+s.size()-1,s.end()); last index
int main()
{
    #ifdef RANA_HOSSAIN
        clock_t tStart = clock();
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    Faster
    test
    {
        int n,i,j,s=1,ans=1,p=3;
        cin>>n;
        if(n==1)
        {
            cout<<"1"<<endl;
            continue;
        }
        while(1)
        {
            if(s+p<=n)
            {
                s+=p;
                p+=2;
                if(s==n)
                    break;
            }
            else
                break;
            //cout<<s<<endl;
            ans++;
        }
        
        cout<<ans+1<<endl;
    }



    #ifdef RANA_HOSSAIN
        fprintf(stderr, "\n>> Runtime: %.10fs\n", (double) (clock() - tStart) / CLOCKS_PER_SEC);
    #endif
}