As we know, if $$$x!=1$$$ and $$$x!=y$$$, then $$$gcd(x,y)<=x/2$$$. That means for each element $$$a_i$$$ in the array, it will change at most $$$log(a_i)$$$ times before it becomes $$$1$$$.

So we can implement the process, but for every round we only consider the element that will be changed (which means $$$a_i != 1$$$ and $$$a_i != a_{i+1}$$$, then we just do $$$O(N*log(a_i))$$$ times change.

How to find the elements that will be changed in the next round? We can see that element will change in the next round, are only in the element changed in this round and their left element, so we just check them. The number of elements we check is $$$O(N*log(a_i))$$$ too.

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<string.h>
#include<set>
#include<map>
#define LL long long
#define pr pair<int,int>
#define fr first
#define sc second
//#define mp make_pair
using namespace std;
LL read( ){LL sum=0;char c=getchar( );bool f=0;while(c<'0' || c>'9') {if(c=='-') f=1;c=getchar( );}while(c>='0' && c<='9') {sum=sum*10+c-'0';c=getchar( );}if(f) return -sum;return sum;}
void read(int &sum){sum=0;char c=getchar( );bool f=0;while(c<'0' || c>'9') {if(c=='-') f=1;c=getchar( );}while(c>='0' && c<='9') {sum=sum*10+c-'0';c=getchar( );}if(f) sum=-sum;}
void read(LL &sum){sum=0;char c=getchar( );bool f=0;while(c<'0' || c>'9') {if(c=='-') f=1;c=getchar( );}while(c>='0' && c<='9') {sum=sum*10+c-'0';c=getchar( );}if(f) sum=-sum;}
const int N=500005;
int n,a[N],b[N],p[N];
int gcd(int x,int y) {return y?gcd(y, x%y):x;}
int main( )
{
  int i,x,y,L,R,T=read( );
  while(T--)
    {
      read(n);
      for(i=1;i<=n;i++) read(a[i]);
      int cnt=0, ans=0;
      for(i=1;i<=n;i++)
	{
	  x=i;
	  y=(x==n?1:x+1);
	  if(a[x]!=a[y]) p[++cnt]=x;
	}
      while(cnt)
	{
	  for(i=1;i<=cnt;i++)
	    {
	      x=p[i];
	      y=(x==n?1:x+1);
	      b[x]=gcd(a[x],a[y]);
	    }
	  set<int>S;
	  for(i=1;i<=cnt;i++)
	    {
	      x=p[i];
	      a[x]=b[x];
	    }
	  for(i=1;i<=cnt;i++)
	    {
	      x=p[i];
	      R=(x==n?1:x+1);
	      L=(x==1?n:x-1);
	      if(a[x]!=a[R] && a[x]!=1) S.insert(x);
	      if(a[L]!=a[x] && a[L]!=1) S.insert(L);
	    }
	  ans++;cnt=0;
	  for(auto &x:S) p[++cnt]=x;
	  if(!cnt) break;
	}
      printf("%d\n",ans);
    }
  return 0;
}