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 lli long long int
#define lp(i,n) for(lli i=0;i<n;i++)
#define mod 1000000007
#define qsort(a,n) sort(a,a+n)
#define pb push_back
#define be begin
#define ft ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fi first 
#define se second
#define ss(x) std::setprecision(x)
#define  pr pair<lli,lli>
#define vec vector<lli>
#define pq priority_queue
#define ub upper_bound
#define lb lower_bound
#define all(v) v.begin(),v.end()
using namespace std;
void init_code()
{
 #ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif // ONLINE_JUDGE
}
                /*Comments
don't declare like this string q=char+char
1.In these cases declare like string q;
q=char+char as here the declare string is already a empty string and thus it is like q=q+char+char 
q=string+char is valid
2. pow(a,n) returns value in m%mod where m=a^n
3.In questions realted to games first try some examples than try either inductive or invariants or greedy kind of approach
_________************_______________*/
const lli MAX=1e6+1;
lli FNI[MAX+1];
lli NNI[MAX+1];
lli fact[MAX+1];
void precall()
{

    NNI[0]=NNI[1]=1;
    for(lli i=2;i<=MAX;i++)
    NNI[i]=NNI[mod%i]*(mod-mod/i)%mod;



    FNI[0]=FNI[1]=1;
    for(lli i=2;i<=MAX;i++)
    FNI[i]=(NNI[i]*FNI[i-1])%mod;


    fact[0]=1;
    for(lli i=1;i<=MAX;i++)
    {
        fact[i]=(fact[i-1]*i)%mod;
    }
}
lli comb(lli n,lli r)
{
    lli ans=(fact[n]*FNI[r])%mod*FNI[n-r]%mod;
    return ans;
}
vector<bool>prime(MAX,1);
void sieve()
{
    for(lli p=2; p*p<MAX;p++)
    {
        if (prime[p] == true)
        {
            for (lli i=p*p; i<MAX;i+=p)
                prime[i] = false;
        }
    }
}
lli solvepow(lli x,lli y)
{
    lli res=1;  
    x=x%mod; 
    if(x==0) return 0;
    while(y>0)
    {
        if(y&1)res=(res*x)%mod;
        y=y>>1;
        x=(x*x)%mod;
    }
    return res;
}
 int main() {
    init_code();
     ft;
   
 lli t;cin>>t;
 while(t--)
 {
    lli s;cin>>s;
    lli r=sqrt(s);
    if(r*r==s)cout<<r<<"\n";
    else cout<<r+1<<"\n";
    
}
}