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 int long long
#define pb push_back
#define ppb pop_back
#define mp make_pair
using pi = pair<int,int>;
using vi = vector<int>;
using vvi = vector<vector<int> >;
#define ff first
#define ss second
#define all(x) x.begin(),x.end()
#define sz(x) (int)(x).size()

template<typename T> istream& operator>>(istream& is,  vector<T> &v){for (auto& i : v) is >> i; return is;}
template<typename T> ostream& operator<<(ostream& os,  vector<T>  v){for (auto& i : v) os << i << ' '; return os;}
template<typename T, typename Y> istream& operator>>(istream& is,  pair<T,Y> &p){ is >> p.first >> p.second; return is;}
template<typename T, typename Y> ostream& operator<<(ostream& os,  pair<T,Y>  p){ os << p.first << ' ' << p.second  << ' '; return os;}

//const int inf = 9e18;
const int NUM=1000030;
const int N = 10000000;
vector<int> lp, sieve;
vector<int> pr;
void primefactor();
int binpow(int a, int b);
int binpow(int a, int b, int mod);
int gcd(int a, int b);
int lcm (int a, int b);
bool comp(int a, int b);
int inversemod(int a, int mod);
void calc_sieve();

void test()
{
    int n=0;
    cin>>n;
    int temp=sqrt(n);
    cout<<temp+(temp*temp==n?0:1)<<endl;
}

signed main()
{
//    freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--) test();
    return 0;
}

//    1. Take care of the corner cases, write one or two tests that seem to be the cases.
//    2. Check the bounds on the intput and decide the algorithm.
//    3. Make the test cases for the bounds, try to include 0, 1, 2 cases in the new test cases, and if possible, scale the cases.
//    4. Check for overflow and the array bounds carefully, always take care of the decreasing loop.

void calc_sieve() //credits: Anish_Sofat
{
    sieve.resize(NUM+1,0);
    for (int x = 2; x <= NUM; x++)
        {
            if (sieve[x]) continue;
            for (int u = x; u <= NUM; u += x)
            {
                    sieve[u] = x ;
            }
        }
}

void primefactor()
{
    lp.resize(N+1,0);
    for (int i=2; i<=N; ++i) {
        if (lp[i] == 0) {
            lp[i] = i;
            pr.push_back (i);
        }
        for (int j=0; j<(int)pr.size() && pr[j]<=lp[i] && i*pr[j]<=N; ++j)
            lp[i * pr[j]] = pr[j];
    }
}

int binpow(int a, int b)
{
    int res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}

int binpow(int a, int b, int mod)
{
    int res = 1;
    while (b > 0) {
        if (b & 1)
            res = (res * a)%mod;
        a = (a * a)%mod;
        b >>= 1;
    }
    return res%mod;
}

int gcd(int a, int b)
{
    if(b==0) return a;
    else return gcd(b,a%b);
}

int lcm (int a, int b)
{
    return ((a / gcd(a, b)) * b);
}

bool comp(int a, int b)
{
    return a>b;
}

int inversemod(int a, int mod)
{
    return binpow(a,mod-2, mod);
}
