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. 
// Problem: A. Find The Array
// Contest: Codeforces - Educational Codeforces Round 111 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1550/problem/A
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

/*
+
++
+++
++++
+++++
++++++
+++++++
++++++++
+++++++++
++++++++++
+++++++++++
++++++++++++
+++++++++++++
++++++++++++++
+++++++++++++++
++++++++++++++++
+++++++++++++++++
++++++++++++++++++
+ +++++++++++++++++
+  ++++++++++++++ ++
+   +++++++++++++  ++
+    ++++++++++++   ++
+     +++++++++++    ++
+      ++++++++++     ++
+       +++++++++      ++ 
+        ++++++++       ++
+         +++++++++++++++++
+          +++++++++++++++++
+           ++++++++++++++
+            +++++++++++
+             ++++++++
+              +++++
+               ++
+               +
+               +
+              ++
+             +++
+            ++++
+           +++++
+           +++++
+           +++++
+           +++++
+     +     +++++
+    +++    +++++
+   ++ ++   +++++
+  ++   ++  +++++
+ ++  +  ++ +++++
+++  +++  +++++++
++  ++ ++  ++++++


 ++++++++      +++++++++++     +++      +++        ++++++++        ++++++++
+++++++++     +++++++++++++    +++      +++       +++    +++      +++    +++
+++          +++   +++   +++   +++      +++      +++   ++++++    +++      +++
+++          +++   +++   +++   +++      +++      +++ +++  +++           +++
+++          +++   +++   +++   +++ ++   +++ ++   ++++++   +++         +++
+++++++++    +++   +++   +++   +++ ++   +++ ++    +++    +++        +++    ++
 ++++++++    +++   +++   +++   +++++    +++++      ++++++++       +++++++++++
*/
#include <stdio.h>
#include <string.h> 
#include <algorithm>
#include <queue>
#define od(x) printf("%d",x)
#define odb(x) printf("%d ",x)
#define odl(x) printf("%d\n",x)
#define odp(x,y) printf("%d %d\n",x,y)
#define ol(x) puts("")
#define old(x) printf("%lld",x)
#define oldb(x) printf("%lld ",x)
#define oldl(x) printf("%lld\n",x)
#define oldp(x,y) printf("%lld %lld\n",x,y)
#define rg(x) for(int i=1;i<=(x);i++){
#define rg_(i,x) for(int i=1;i<=(x);i++){
#define gr }
#define rrg(x) for(int i=0;i<(x);i++)
#define rdln(a) a[i]=read();
#define rdln0(a,x) rrg(x) rdln(a) gr
#define rdln1(a,x) rg(x) rdln(a) gr
//#define int long long
#define newe(n) struct Edge{int v,w,nxt;}e[n*2+5];\
typedef int arr[n+5];\
arr h;\
int cnt=1;\
inline void addedge(int u,int v,int w){e[cnt]=(Edge){v,w,h[u]};h[u]=cnt++;}
#define mgs int fa[1<<22],sz[1<<22],t[1<<22];\
inline int f(int x){return x==fa[x]?x:fa[x]=f(fa[x]);}\
inline int u(int x,int y)\
{\
	int fx=f(x),fy=f(y);\
	if(fx==fy)return 0;\
	if(sz[fx]>sz[fy])fx^=fy^=fx^=fy;\
	fa[fx]=fy,sz[fy]+=sz[fx];\
	return 1;\
}
inline int read()
{
	int num=0,f=1;char c=getchar();
	while(c<48||c>57){if(c=='-')f=-1;c=getchar();}
	while(c>47&&c<58)num=num*10+(c^48),c=getchar();
	return num*f;
}
inline int re1d()
{
	char c=getchar();
	while(c<48||c>49)c=getchar();
	return c&1;
}
inline int min(int a,int b){return a>b?b:a;}
inline int max(int a,int b){return a<b?b:a;}
signed main()
{
	int T=read();
	while(T--)
	{
		int n=read();
		int c=0,f=0;
		for(int i=1;;i+=2)
		{
			c++;
			f+=i;
			if(f>=n)break;
		}
		odl(c);
	}
	return 0;
}