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 M 1000000007
#define U 998244353
#define N 1000005
#define int long long
#define sz(c) (int)c.size()
#define fr first
#define ll long long 
#define sc second
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()
#define rep(i,a,n) for(int i=a ; i<n ; i++)
#define r0 return 0;
#define endl '\n'
#define INF (int)1e15

/* Debug Begins */ 
# define trace(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); \
  stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
string to_string(char c) { return "'" + string(1, c) + "'";}
string to_string(string s) { return '"' + s + '"';}
string to_string(bool f) { if(f) return "True"; else return "False";}
string to_string(const char* s) { return to_string((string) s);}
template<typename A> string to_string(A);
template<typename A, typename B> string to_string(pair<A, B> p){
  return "(" + to_string(p.first) + ": " + to_string(p.second) + ")";}
template<typename A> string to_string(A v) {bool f = false; string r = "{"; 
  for (auto x: v) {if (f) r += ", "; r += to_string(x); f = true;} return r += "}";}
template<typename A> string to_string(vector<vector<A>> v) {string r; 
  for (auto x: v) r += "\n" + to_string(x); return r;}
int Nerr;
template<typename A> string to_string(A *p) {return to_string(vector<A>(p, p + Nerr));}
void err(istream_iterator<string>) { cerr << endl; }
template<typename T,typename... Args> void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << to_string(a) << "; "; err(++it, args...); }
template<typename T> void kek(T ans) {cout << ans << endl; exit(0);}
#define Lu(...) [&] (auto &&u) { return __VA_ARGS__; }
#define Luv(...) [&] (auto &&u, auto &&v) { return __VA_ARGS__; }
/***************************************************************/



signed main()
{
    ios_base::sync_with_stdio(0);
    int TESTS=1;
    cin>>TESTS;
    while(TESTS--)
    {   
        int s;
        cin >> s;
        rep(n,1,1000){
            if(s >= n and s <= n*n){
                cout << n << endl;
                break;
            }
        }
    }
}