題解：

定義dp狀態f[N] [gcd(a[i-2],a[i-1],a[i])] [gcd(a[i-1],a[i])] [a[i]]。但是很多狀態是沒有用到的，實際上對於m=100時後面三維加起來也不會超過2000，a[i]是gcd(a[i-1],a[i])整數倍，gcd(a[i-1],a[i])是gcd(a[i-2],a[i-1],a[i])的整數倍。

狀態轉移時列舉新的一維，轉移方程為：

f[N+1][ gcd(a[i-1],a[i],a[i+1])) ] [ gcd(a[i],a[i+1]) ] [ a[i+1] ] += f[N] [gcd(a[i-2],a[i-1],a[i])] [gcd(a[i-1],a[i])] [a[i]] * v[ gcd(a[i-2],a[i-1],a[i],a[i+1] ]。

最後答案除以總的概率的逆元。

方程很長，可以預處理狀態壓縮一下。

#include"bits/stdc++.h"
using namespace std;
typedef long long LL;
int T,n,m;
const int mod = 1e9+7;
const int N = 105;
const int M = 4005;
int a[N],v[N],gcd[N][N],id[N][N][N];
int g[M][N],w[M][N];
LL f[N][M];

LL qpow(LL a, LL n)
{
    LL ret = 1;
    while(n){
        if(n&1) ret = ret*a%mod;
        a = a*a%mod;
        n >>= 1;
    }
    return ret;
}

void solve()
{
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++)
        scanf("%d",&a[i]);
    for(int i = 1; i <= m; i++)
        scanf("%d",&v[i]);
    for(int i = 1; i <= m; i++)
        for(int j = 1; j <= m; j++)
            gcd[i][j] = __gcd(i,j);

    int cnt = 0;
    for(int i = 1; i <= m; i++)
        for(int j = i; j <= m; j+=i)
            for(int k = j; k <= m; k+=j)
                id[i][j][k] = ++cnt;  //後面三維壓縮狀態

    for(int i = 1; i <= m; i++)
        for(int j = i; j <= m; j+=i)
            for(int k = j; k <= m; k+=j)
            {
                int x = id[i][j][k];
                for(int y = 1; y <= m; y++)
                {
                    g[x][y] = id[gcd[j][y]][gcd[k][y]][y];  //轉移時的下一個狀態
                    w[x][y] = v[gcd[i][y]];   //轉移時要乘上的權值
                }
            }
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= cnt; j++)
            f[i][j] = 0;
    for(int i = 1; i <= m; i++)
        for(int j = 1; j <= m; j++)
            for(int k = 1; k <= m; k++)
            {
                if(a[1] && a[1] != i)
                    continue;
                if(a[2] && a[2] != j)
                    continue;
                if(a[3] && a[3] != k)
                    continue;
                f[3][ id[gcd[i][gcd[j][k]]][gcd[j][k]][k] ] += 1;
            }
    for(int i = 3; i < n; i++)
        for(int j = 1; j <= cnt; j++)
            for(int k = 1; k <= m; k++)
            {
                if(a[i+1] && a[i+1] != k) continue;
                int nxt = g[j][k];
                f[i+1][nxt] = (f[i+1][nxt]+f[i][j]*w[j][k])%mod;
            }
    LL ans = 0;
    for(int j = 1; j <= cnt; j++)
        ans = (ans+f[n][j])%mod;
    for(int i = 1; i <= n; i++)
        if(!a[i])
            ans = ans*qpow(m,mod-2)%mod;
    cout<<ans<<endl;
}

int main()
{
#ifdef LOCAL
    freopen("input.txt","r",stdin);
#endif // LOCAL
    for(scanf("%d",&T); T; T--)
        solve();
    return 0;
}