题目

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
#define ll long long

const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5, base = 37;
const int N = 1e4;
// const int mod = 1e9 + 7;
// const int mod = 998244353;
const __int128 mod = 212370440130137957LL;

int n, m;
int a[maxn], b[maxn];
//long long ? maxn ? n? m?


void solve(){
    ll res = 0;
    int x;
    cin >> n >> x;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    vector<int> divs;
    divs.pb(0);
    vector<int> id(x + 1, 0);
    for(int i = 1; i * i <= x; i++){
        if(x % i == 0){
            divs.pb(i);
            // id[i] = divs.size() - 1;
            if(x / i != i){
                divs.pb(x / i);
                // id[x / i] = divs.size() - 1;
            }
        }
    }
    int sz = divs.size() - 1;
    sort(divs.begin() + 1, divs.begin() + sz + 1);//排序因为后面要从大的因数开始更新
    for(int i = 1; i <= sz; i++){
        id[divs[i]] = i;//x的因数到编号的映射
    }
    vector<int> can(sz + 1, 0);//can[i]表示因数divs[i]能不能被凑出
    can[1] = 1;
    res = 1;
    for(int i = 1; i <= n; i++){
        if(x % a[i] != 0) continue;//不是x的因数，直接忽略
        if(can[id[x / a[i]]]){
            res++;
            can.assign(sz + 1, 0);
            can[1] = 1;
            can[id[a[i]]] = 1;
            continue;
        }
        //必须从大的因数开始更新can，
        //设
        //下标 : 1 2 3 4
        //divs : 1 2 4 8
        //can  : 1 0 0 0
        //当前数a[i] : 2
        //若从小的因数开始更新，那么更新之后
        //can  : 1 1 1 1 （显然错误）
        for(int j = sz; j >= 1; j--){
            if(divs[j] % a[i] == 0 && can[id[divs[j] / a[i]]]){
                can[j] = 1;
            }
        }
    }
    cout << res << '\n';
}
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);

    cout << fixed << setprecision(9);

    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}