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 v vector<ll>
#define vv vector<v>
#define pii pair<ll,ll>
#define f first
#define s second
#define pb push_back
#define endl "\n"
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define all(arr) arr.begin()+1,arr.end()
ll mod1 = 1000000007;
ll mod2 = 998244353;
ll inf = mod1*mod1;
ll powe(ll a,ll b)
{
    ll res=1;
    while(b)
    {
        if(b%2==1)
            res=(res*a)%mod1;
        b/=2;
        a=(a*a)%mod1;
    }
    return res;
}
void solve()
{
    ll n,i,sum,ans;
    cin>>n;
    if(n==1)
    {
        cout<<1<<endl;
        return;
    }
    sum=1,ans=1;
    for(i=3;i<=10000;i+=2)
    {
        if(i+sum<n)
            sum+=i,ans++;
        else if(i+sum==n)
        {
            sum+=i;
            ans++;
            break;
        }
        else
        {
            ans++;
            break;
        }
    }
    cout<<ans<<endl;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r",stdin);
    freopen("output.txt", "w",stdout);
#endif
    int t=1;
    cin>>t;
    while(t--)
        solve();
}