#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<math.h>
using namespace std;
#define maxn 10005
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
//每个青蛙，可以跳到gcd(m,a[i])*k的位置
int ppp[maxn];
int num[maxn],vis[maxn];
int main()
{
    int tt;scanf("%d",&tt);
    for(int cas=1;cas<=tt;cas++)
    {
        int n,m;
        int cnt = 0;
        memset(vis,0,sizeof(vis));
        memset(num,0,sizeof(num));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=sqrt(m);i++)//把因子全部筛出来
        {
            if(m%i==0)
            {
                ppp[cnt++]=i;
                if(i*i!=m)
                    ppp[cnt++]=m/i;
            }
        }
        sort(ppp,ppp+cnt);
        for(int i=0;i<n;i++)
        {
            int x;scanf("%d",&x);
            int kk = gcd(x,m);
            for(int j=0;j<cnt;j++)
                if(ppp[j]%kk==0)//说明这个因子的所有，都是可以被跳到的位置
                    vis[j]=1;
        }
        vis[cnt-1]=0;//显然 m是不可能被跳到的
        long long ans = 0;
        for(int i = 0; i < cnt; i++)
        {
            if(vis[i] != num[i])
            {
                int t = (m-1)/ppp[i];
                ans += (long long)t*(t+1)/2 * ppp[i] * (vis[i]-num[i]);
                //容斥一波
                //一开始vis[i] - num[i] = 1的
                //对于每个因数，如果重复计算了，在之后，减去就好了
                t = vis[i] - num[i];
                for(int j = i; j < cnt; j++)
                    if(ppp[j]%ppp[i] == 0)
                        num[j] += t;
            }
        }
        printf("Case #%d: %lld\n",cas,ans);
    }
}

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n")
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl
#define ffread(a) fastIO::read(a)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=1e4+10;
ll gcd(ll a,ll b)
{
    return b?gcd(b,a%b):a;
}
ll sum(ll x,ll n)
{
    ll temp=(n-1)/x;
    return temp*x+(temp*(temp-1)/2)*x;
}
int main()
{
    ll t,n,m;
    int kase=1;
    scanf("%lld",&t);
    while(t--)
    {
        set<ll> s;
        scanf("%lld%lld",&n,&m);
        for(int i=0;i<n;i++)
        {
            ll x;
            scanf("%lld",&x);
            s.insert(gcd(x,m));
        }
        vector<ll> g;
        for(auto it:s)
        {
            int flag=1;
            for(auto itt:s)
                if(it%itt==0&&it!=itt)
                 flag=0;
            if(flag)g.pb(it);
        }
        ll cnt=g.size();
        ll ans=0;
        for(ll i=0; i<(1ll<<cnt); i++)
        {
            ll temp=1,jishu=0;
            for(ll j=0; j<cnt; j++)
            {
                if(i&(1<<j))
                    temp=temp*g[j]/gcd(g[j],temp),jishu++;
            }
            if(jishu==0)
                continue;
            if(jishu%2==1)
                ans+=sum(temp,m);
            else
                ans-=sum(temp,m);
        }
        printf("Case #%d: %lld\n",kase++,ans);
    }
}