/*
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
", ans);
}
}