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;
inline int read()
{
    int x=0,f=1;
    char ch = 0;
    while(!isdigit(ch)) {if(ch=='-') f=-1; ch = getchar();}
    while(isdigit(ch)) {x = x*10+ch-'0'; ch = getchar();}
    return x*f;
}
typedef long long ll;
const int maxn = 2e5+100;
const double eps = 1e-13;
int T,n;
/*
double c,m,p,v,_v;
double ans;
map<pair<double,double>,double> mp;
double dfs(double a,double b, double c,double nw,int tt)
{
    //cout << a << ' ' << b << ' ' << c << endl;
    if(mp.find(make_pair(a,b))!=mp.end())
    {
        ans += mp[make_pair(a,b)];
        return mp[make_pair(a,b)];
    }
    double hv = c*nw;
    if(a>v)
    {
        if(b>eps) hv += dfs(a-v,b+_v,c+_v,nw*a,tt+1);
        else hv += dfs(a-v,0,c+v,nw*a,tt+1);
    }
    else if(a>eps)
    {
        if(b>eps) hv += dfs(0,b+a/2,c+a/2,nw*a,tt+1);
        else hv += a*nw*(tt+1);
    }
    if(b>v)
    {
        if(a>eps) hv += dfs(a+_v,b-v,c+_v,nw*b,tt+1);
        else hv += dfs(0,b-v,c+v,nw*b,tt+1);
    }
    else if(b>eps)
    {
        if(a>eps) hv += dfs(a+b/2,0,c+b/2,nw*b,tt+1);
        else hv += b*nw*(tt+1);
    }
    ans += tt*hv;
    return mp[make_pair(a,b)] = mp[make_pair(b,a)] = hv;
}
int main()
{
    T = read();
    while(T--)
    {
        scanf("%lf%lf%lf%lf",&c,&m,&p,&v);
        _v = v/2;
        mp.clear();
        ans = 0;
        printf("%.13f\n",dfs(c,m,p,1));
        cout << "---" << ans << endl;
    }
    return 0;
}
*/
int main()
{
    T = read();
    int s;
    while(T--)
    {//(1+x*2-1)*x / 2 = s
        s = read();
        int x = sqrt(s);
        printf("%d\n",x+(x*x!=s));
    }
}
