#include<bits/stdc++.h>
#define int long long
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define pb push_back
#define pii pair<int,int>
using namespace std;
//const int mod=998244353;
int qpw(int a,int b,int mod){int ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
//int inv(int x){return qpw(x,mod-2);}
const int MAX_N = 5e4+10;
const int LOG = 17;  // log2(MAX_N) 的上界
int n, a[MAX_N];
int f[MAX_N][LOG];   // 倍增表：up[i][k] 表示从 i 向上跳 2^k 步后的节点
int gcd_table[MAX_N][LOG];  // gcd_table[i][k] 表示从 i 到 up[i][k] 的 GCD 值
int parent[MAX_N];    // 笛卡尔树中的父节点
int depth[MAX_N];     // 每个节点的深度

int dp_gcd[17][(int)5e4 + 1];
// 求 GCD
int gcd(int a, int b) {
    return b?gcd(b,a%b):a;
}
void solve(){
    int k;cin>>n>>k;
    set<int>sb;
    vector<int>ls_min(n+1),rs_min(n+1);
    int root_min;
    for(int i=1,x;i<=n;i++)cin>>a[i],sb.insert(a[i]);
    if(sb.size()==1){
        cout<<k<<" "<<k*(k+1)/2<<'\n';
        return;
    }
    vector<int>sta(n+1);
    function<void()>build_min=[&](){
        int top=0,cur=0;
        for(int i=1;i<=n;i++){
            cur=top;
            while(cur&&a[sta[cur]]>a[i])cur--;
            if(cur)rs_min[sta[cur]]=i;
            if(cur<top)ls_min[i]=sta[cur+1];
            sta[++cur]=i;
            top=cur;
        }
    };
    auto query_min=[&](int l,int r){
        int x=root_min;
        for(;;x=(r<x?ls_min[x]:rs_min[x]))if(l<=x&&x<=r)return x;
    };
    int gc=0;
    vector<int>b(n+1);
    for(int i=2;i<=n;i++){
        gc=gcd(gc,b[i]=abs(a[i]-a[i-1]));
    }
    build_min();
    root_min=sta[1];
    set<int>dp,ndp;
//    cout<<query(2,2)<<'\n';
    function<int(int,int)> dfs = [&]( int l, int r)->int
    {
        if (l == r)
        return 0ll;
        int mid = query_min(l, r);
        if(l==1&&r==n) {
            for (int i = 1; i * i <= gc; i++) {
                if (gc % i)
                    continue;
                for (auto j: (i * i == gc ? vector<int>{i} : vector<int>{i, gc / i})) {
                    {
                        if (j > a[mid] and j - a[mid] <= k)
                            dp.insert(j - a[mid]);
                    }
                }

            }
            int a=dfs(l, mid - 1);
            int b=dfs(mid + 1, r);
            return gc;
        }else {
               int L=dfs(l,mid-1);
               int R=dfs(mid+1,r);
               int G=gcd(L,R);
               G=gcd(G,b[mid+1]);
               for(int x:dp){
//                   cout<<x<<" ";
                   if(G%(x+a[mid]))continue;
                   ndp.insert(x);
               }
//               cout<<'\n';
               dp.swap(ndp);
               ndp.clear();
               return G;
        }
    };
    int ooo=dfs(1, n);
    int sum=0;
    for(int x:dp)sum+=x;
    cout<<dp.size()<<" "<<sum<<'\n';
}
signed main(){
    ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);
    int _=1;
    cin>>_;
    while(_--)solve();
}
/*
3
1
3 1000000000
230 286458 41263680

 1
10 1000000000
230 286458 41263680 41263680 41263680 41263680 41263680 41263680 41263680 41263680 41263680

 */