/*

HDU 6073 - Matching In Multiplication [ 图论 ]  |  2017 Multi-University Training Contest 4

题意：

	定义一张二分图，U中每个节点和V中两个节点连边

	完美匹配的权值为该匹配所有边的权值相乘

	求所有完美匹配的权值之和

分析：

	可以发现有些V中的点只能连唯一的U中的点

	按拓扑排序思路将这些全部处理掉后，剩下的点构成一个个环

	每个环有两种连线方式，间隔取边权

*/

#include <bits/stdc++.h>

using namespace std;

#define LL long long

const int N = 600005;

const LL MOD = 998244353;

struct Edge {

    int v; LL w;

};

vector<Edge> G[N];

int vis[N];

int cnt[N];

int t, n;

LL ans;

queue<int> que;

LL s[2];

void dfs(int x, int pre, int p)

{

    if (vis[x]) return;

    vis[x] = 2;

    for (const auto & e : G[x])

    {

        if (vis[e.v] == 1 || e.v == pre) continue;

        s[p] = s[p] * e.w % MOD;

        dfs(e.v, x, p^1);

        break;

    }

}

void solve()

{

    for (int i = n+1; i <= 2*n; i++)

        if (G[i].size() == 1) que.push(i);

    ans = 1;

    while (!que.empty())

    {

        int y = que.front(); que.pop();

        vis[y] = 1;

        for (const auto& e: G[y])

        {

            if (!vis[e.v])

            {

                vis[e.v] = 1;

                ans = ans * e.w % MOD;

                for (const auto & ee: G[e.v])

                {

                    if (vis[ee.v]) continue;

                    cnt[ee.v]--;

                    if (cnt[ee.v] == 1) que.push(ee.v);

                }

            }

        }

    }

    for (int i = 1; i <= n; i++)

    {

        if (!vis[i])

        {

            s[0] = s[1] = 1;

            dfs(i, i, 0);

            ans = ans * (s[0] + s[1]) % MOD;

        }

    }

}

void init(int n)

{

    for (int i = 0; i <= n; i++) G[i].clear();

    while (!que.empty()) que.pop();

    memset(vis, 0, sizeof(vis));

    memset(cnt, 0, sizeof(cnt));

}

int main()

{

    scanf("%d", &t);

    while (t--)

    {

        scanf("%d", &n);

        init(n<<1);

        for (int i = 1; i <= n; i++)

        {

            int v; LL w;

            scanf("%d%lld", &v, &w); v += n;

            G[i].push_back(Edge{v, w});

            G[v].push_back(Edge{i, w});

            cnt[v]++;

            scanf("%d%lld", &v, &w); v += n;

            G[i].push_back(Edge{v, w});

            G[v].push_back(Edge{i, w});

            cnt[v]++;

        }

        solve();

        printf("%lld\n", ans);

    }

}