Description
There is a complete graph containing \(n\) vertices, the weight of the \(i\)-th vertex is \(w_i\).
The length of edge between vertex \(i\) and \(j\) \((i≠j)\) is \(\lfloor \sqrt{|w_i-w_j|}\rfloor\).
Calculate the length of the shortest path from 1 to \(n\).

Input
The first line of the input contains an integer \(T (1\leqslant T\leqslant 10)\) denoting the number of test cases.
Each test case starts with an integer \(n (1\leqslant n\leqslant 10^5)\) denoting the number of vertices in the graph.
The second line contains n integers, the \(i\)-th integer denotes \(w_i (1\leqslant w_i\leqslant 10^5)\).

Output
For each test case, print an integer denoting the length of the shortest path from 1 to \(n\).

Sample Input

1
3
1 3 5

Sample Output

2

---

根据\(\lfloor\sqrt{a}\rfloor+\lfloor\sqrt{b}\rfloor\geqslant\lfloor\sqrt{a+b}\rfloor,(a,b\in\Z)\)可得，对于任意两点\(i,j\)而言，必然有\(\lfloor\sqrt{|w_i-w_k|}\rfloor+\lfloor\sqrt{|w_k-w_j|}\rfloor\geqslant\lfloor\sqrt{|w_i-w_k|+|w_k-w_j|}\rfloor\geqslant\lfloor\sqrt{w_i-w_j}\rfloor\)，故我们直接计算\(1\sim n\)即可

现对\(\lfloor\sqrt{a}\rfloor+\lfloor\sqrt{b}\rfloor\geqslant\lfloor\sqrt{a+b}\rfloor\)进行证明

令\(m^2\leqslant a\leqslant (m+1)^2\)和\(n^2\leqslant b\leqslant (n+1)^2\)，则有\(\lfloor\sqrt{a}\rfloor=m,\lfloor\sqrt{b}\rfloor=n\)

若\(m=0\)或\(n=0\)，则\(a=0\)或\(b=0\)，此时命题显然成立

欲证 \(m+n\geqslant\lfloor\sqrt{a+b}\rfloor\)

则只需证 \(m+n+1\geqslant \sqrt{a+b}\)

则只需证 \(m+n+1\geqslant\sqrt{(m+1)^2+(n+1)^2}\)

则只需证 \(m^2+n^2+1+2m+2n+2mn\geqslant m^2+1+2m+n^2+1+2n\)

则只需证 \(2nm>1\)，这在\(n,m>0\)时是显然成立的，故得证

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e5;
int A[N+10];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read(0);
	while (T--){
		int n=read(0);
		for (int i=1;i<=n;i++)	A[i]=read(0);
		printf("%d\n",(int)trunc(sqrt(abs(A[n]-A[1]))));
	}
	return 0;
}