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

using ll = long long;
#define endl '\n'

const int MOD = 1e9 + 7;
const int N = 5e5 + 5;
int a[N], b[N], w[N];
vector<bool> f(N);
int ans[N], fac[N];
int n, cnt;

ll qpow(ll a, ll b, ll p)
{
    ll tmp = 1;
    a = a % p;
    while (b)
    {
        if (1 & b)
            tmp = tmp * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return tmp % p;
}

ll inv(ll a, ll p) // 费马小定理求逆元
{
    return qpow(a, p - 2, p);
}

int dfs(int u)
{
    int v = b[u];
    if (u == v)
    {
        ans[u] = a[u] % MOD;
        return 0;
    }
    // cout<<u<<" "<<v<<endl;
    if (a[u] < a[v])
    {
        ans[u] = ((a[u] + w[u]) % MOD + MOD) % MOD;
        f[u] = 1;
        return 1;
    }
    else if (a[u] >= a[v] + w[v])
    {
        ans[u] = a[u];
        f[u] = 1;
        return 0;
    }
    else if (a[u] >= a[v])
    {

        int ok = dfs(v);
        cnt++;
        if (ok == 1)
        {
            ll x = fac[cnt];
            ll y = fac[n];
            x = ((1ll * y * inv(x, MOD) % MOD) + MOD) % MOD;
            ll xx = ((1ll * a[u] * (y - x) % MOD + (w[u] + a[u]) * x % MOD) + MOD) % MOD;
            ll yy = y;
            ans[u] = (1ll * xx * inv(yy, MOD) % MOD + MOD) % MOD;
            f[u] = 1;
            return 1;
        }
        else
        {
            ans[u] = a[u] % MOD;
            f[u] = 1;
            return 0;
        }
    }
    return 0;
}
void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> b[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> w[i];
        f[i] = 0;
    }

    for (int i = 1; i <= n; i++)
    {
        cnt = 1;
        if (f[i] == 0)
            dfs(i);
    }

    for (int i = 1; i <= n; i++)
    {
        cout << (ans[i] % MOD + MOD) % MOD << " ";
    }
    cout << endl;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    fac[0] = 1;
    for (int i = 1; i <= N; ++i)
        fac[i] = ((1ll * fac[i - 1] * i) % MOD + MOD) % MOD;

    int t = 1;
    cin >> t;

    while (t--)
    {
        solve();
    }
}