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>
using namespace std;
#define FAST ios_base::sync_with_stdio(0),cin.tie(NULL),cout.tie(NULL)
typedef pair<long long int,long long int> pll;
#define ll long long int
#define ld long double
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define sz size()
#define input(a) for(auto &x:a){cin>>x;}
#define all(v) v.begin(),v.end()
#define precise(x) fixed<<setprecision(x)
#define MOD 1000000007
const int N = 2e5 + 5;
const long long int MOOD = 998244353 ;

bool sortbysec(const pair<ll,ll> &a,const pair<ll,ll> &b)
{
    return(a.second<b.second);
}
ll power(ll a,ll b)
{
    ll c=1;
    while(b)
    {
        if(b&1)
        {
            //c=((ll)c*a)%MOD;
            c=((ll)c*a);
        }
        //a=((ll)a*a)%MOD;
        a=((ll)a*a);
        b=b/2;
    }
    return c;
}
ll fact(ll n)
{
    if(n==0)
        return 1;
    return (n*fact(n-1));
}

void solve()
{
    ll n;
    cin>>n;
    ll ans =0;
    ll cnt=0,i=0;
    for(i=1;;i+=2)
    {
        if(i>=n)
        {
            break;
        }
        n-=i;
        ans++;
    }
    if(n>0)
        ans++;
    cout<<ans<<endl;
    return;
}

int main()
{
    FAST;
    ll t;
    t = 1;
    cin>>t;
    while(t--)
    {
        solve();
    }
}
