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 <iostream>
#include <random>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <set>
#include <string>
#include <queue>
#include <bitset>
#include <map>
#include <ctime>
#include <unordered_set>
//#include <bits/extc++.h>
#include <unordered_map>

#pragma GCC optimize("O3")
//#pragma GCC target("sse,sse2,sse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
//#pragma GCC optimization ("unroll-loops")
//#define sz(x) signed(x.size())
#define db(x) cout << #x << " " << x << endl
#define int long long
using namespace std;
//using namespace __gnu_cxx;
//using namespace __gnu_pbds;
//using ordered_set = tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>;

//mt19937 rnd(505);
const int INF = 2000000007;
const int M = 1000000007;
const int N = 1e5 + 7;
const int k = 63;
//const int p = 37;

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout.precision(37);
//    freopen("friends.in", "r", stdin);
//    freopen("friends.out", "w", stdout);
    cout << fixed;
//    string s;
//    cin >> s;
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
//        cout << n << endl;
        int summ = 1;
        if (n == 1) {
            cout << 1 << endl;
            continue;
        }
        for (int i = 0; i < n; ++i) {
            if (summ + (1 + i) * 2 + 1 >= n) {
                cout << i + 2 << endl;
                break;
            }
            summ += (1 + i) * 2 + 1;
        }
    }
}