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 ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define f(i,m,n) for(int i = m; i < n; i++)
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vl vectaor<ll>
#define vii vector<pii>
#define vll vector<pll>
#define vvi vector<vi>
#define vvl vector<vl>
#define ff first
#define ss second
#define FAST ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);srand(time(NULL));
#define INF LONG_LONG_MAX
#define MINF LONG_LONG_MIN

const int MOD=1000000007;



 ll  power(ll x,ll  n)
 {
      ll res=1;
     while(n)
     {
         if(n&1)
         {
             res=(res*x);
         }
         n>>=1;
         if(n!=0)
         {
              x=(x*x);
         }
        
     }
     return res;
 }
 
 ll  powerMOD(ll x,ll  n)
 {
      ll res=1;
     while(n)
     {
         if(n&1)
         {
             res=(res*x)%MOD;
         }
         n>>=1;
         x=(x*x)%MOD;
     }
     return res;
 }
 

bool mycompare(pair<ll,ll> &a,pair<ll,ll> &b)
{
    if(a.ff==b.ff)
    {
        return a.ss<b.ss;
    }
    return a.ff>b.ff;
}


bool mycompare1(pair<int,int> &a,pair<int,int> &b)
{
    return a.ss<b.ss;
}

ll go(ll &n,ll &x) {
    ll m=0;
    while(x<=n) {
        n-=x;
        x+=2;
        m++;
    }
    return m;
}

void solve()
{
    ll i,j,k,n,m,l,x;
    cin>>n;
    m=0;
    x=1;
    m+=go(n,x);
    while(n) {
        if(x<=n) {
            n-=x;
            m++;
        } else {
            x--;
        }
    }
    cout<<m<<"\n";
}

int main()
{
    FAST;
    int t=1,i;
    cin>>t;
    f(i,0,t) solve();
    return 0;
}