/*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 
     
      using namespace std;#define LL long longconst int N = 600005;const LL MOD = 998244353;struct Edge {    int v; LL w;};vector
      
        G[N];int vis[N];int cnt[N];int t, n;LL ans;queue
       
         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);    }}