RT，哪里假了awa，思路是分解质因数，考虑将$w$质因子放到$p_i$中的贡献

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 1e4 + 10;
const int mod = 998244353;
LL n, w;
LL a[N];

LL p[N][100], c[N][100], tot;
LL sum[N];

void divide(int x)
{
    tot ++;
    int cnt = 0;
    for (int i = 2; i * i <= x; i ++ )
        if (x % i == 0)
        {
            p[tot][++ cnt] = i, c[tot][cnt] = 0;
            while (x % i == 0)
            {
                x /= i;
                c[tot][cnt] ++;
            }
        }
    if (x > 1)
        p[tot][++ cnt] = x, c[tot][cnt] = 1;

    p[tot][0] = cnt;
    sum[tot] = 1; // 每个数因数个数
    for (int i = 1; i <= cnt; i ++ )
        sum[tot] = sum[tot] * (c[tot][i] + 1);
    /*
    for (int i = 1; i <= cnt; i ++ )
        cout << p[tot][i] << '^' << c[tot][i] << endl;
    cout << endl;*/
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> w;
    divide(w);
    for (int i = 1; i <= n; i ++ )
    {
        cin >> a[i];
        divide(a[i]);
    }
    /*
    for (int i = 1; i <= tot; i ++ )
        cout << sum[i] << ' ';*/

    LL ans = 1;
    for (int i = 1; i <= p[1][0]; i ++ ) // w各个因数
    {
        int v = p[1][i];
        int cnt = 0;
        for (int j = 2; j <= tot; j ++ )
            for (int k = 1; k <= p[j][0]; k ++ )
                if (p[j][k] == v) cnt ++;
        if (cnt == n) ans = ans * c[1][i] % mod;
        else ans = ans * (c[1][i] + 1) % mod;
    }

    
    for (int i = 2; i <= tot; i ++ )
    {
        ans = ans * sum[i] % mod;
        //cout << sum[i] << endl;
    }
    cout << ans << endl; 

    return 0;
}