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>
/*
#include<iostream>
#include<vector>
#include<queue>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string>
#include<string.h>
#include<map>
#include<math.h>*/
using namespace std;
const int lmt  = 100005;
const long long MOD = 1000000007;
const long long mod = 998244353;
const int AbcN = 'z'-'a';

#define f first 
#define S second
#define ll long long
#define ld long double
#define ull unsigned long long
#define forn(i, n) for(int i = 0; i < int(n); i++)
#define for1(i, n) for(int i = 1; i < int(n); i++)
#define forv(i,a,n) for(int i = int(a); i < int(n); i++)
#define rof(i, n) for(int i = int(n); i >= 0; i--)
#define rofv(i,a,n) for(int i = int(n); i >= int(a); i--)
#define pb push_back
#define ins insert
#define pai pair<int, int>
#define pal pair<long long, long long>
#define vi vector<int>
#define vl vector<long long>
#define vld vector<long double>
/*
struct wer{
    int a, b, c;
};
bool operator < (const wer &a, const wer &b){
    return a.a < b.a;
}*/
ll gcd(ll a, ll b){ return (b==0 ? a : gcd(b, a % b)); }
void solve(){
    int n; cin>>n;
    int c = 1, d = 1, i = 3;
    while(c<n){
        c+=i;
        i+=2;
        d++;
    }
    cout<<d<<"\n";
    return;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int q=1;
    cin>>q;
    while(q--){
        solve();
    }
    return 0;
}