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<cstdio>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<vector>
#include<map>
#include<algorithm>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<set>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;++i)
#define inf 0x3f3f3f3f;
typedef pair<int,int> PII;
typedef long long ll;
inline int lowbit(int x){//返回x二进制中最后一个1对应的十进制数字 
	return x&(-x);
}
inline int gcd(int x,int y){//最大公约数
	int z = y;
	while (x % y != 0)
	{
		z = x % y;
		x = y;
		y = z;
	}
	return z;
}
inline int lcm(int x,int y){//最小公倍数
	ll z = x * y;
	return z / gcd(x,y);
}
/*---------------------------------分界线------------------------------------------*/

const int N = 1e6+10;
ll t;

int main(){
	cin>>t;
	while(t--){
		int n;
		scanf("%d",&n);
		
		if(n == 1) cout<<1<<endl;
		else if(n == 2 || n == 3) cout<<2<<endl;
		else{
			int cnt1=2,cnt2=1;
			int sum1=3,sum2=1;
			int x1 = 2;
			while(sum1 < n){
				x1+=2;
				cnt1++;
				sum1+=x1;
				// cout<<x<<" ";
			}
			// cout<<endl;
			
			int x2=1;
			while(sum2 < n){
				x2+=2;
				cnt2++;
				sum2+=x2;
			}
			// cout<<cnt1<<endl<<cnt2<<endl;
			cout<<min(cnt1,cnt2)<<endl;
		}

	}
}