Problem: Given are simple undirected graphs X, Y, Z, with N vertices each and M_1, M_2, M_3 edges, respectively. The vertices in X, Y, Z are respectively called x_1, x_2, \dots, x_N, y_1, y_2, \dots, y_N, z_1, z_2, \dots, z_N. The edges in X, Y, Z are respectively (x_{a_i}, x_{b_i}), (y_{c_i}, y_{d_i}), (z_{e_i}, z_{f_i}).

Based on X, Y, Z, we will build another undirected graph W with N^3 vertices. There are N^3 ways to choose a vertex from each of the graphs X, Y, Z. Each of these choices corresponds to the vertices in W one-to-one. Let (x_i, y_j, z_k) denote the vertex in W corresponding to the choice of x_i, y_j, z_k.

We will span edges in W as follows:

* For each edge (x_u, x_v) in X and each w, l, span an edge between (x_u, y_w, z_l) and (x_v, y_w, z_l).
* For each edge (y_u, y_v) in Y and each w, l, span an edge between (x_w, y_u, z_l) and (x_w, y_v, z_l).
* For each edge (z_u, z_v) in Z and each w, l, span an edge between (x_w, y_l, z_u) and (x_w, y_l, z_v).



Then, let the weight of the vertex (x_i, y_j, z_k) in W be 1,000,000,000,000,000,000^{(i +j + k)} = 10^{18(i + j + k)}. Find the maximum possible total weight of the vertices in an independent set in W, and print that total weight modulo 998,244,353.

Constraints

* 2 \leq N \leq 100,000
* 1 \leq M_1, M_2, M_3 \leq 100,000
* 1 \leq a_i, b_i, c_i, d_i, e_i, f_i \leq N
* X, Y, and Z are simple, that is, they have no self-loops and no multiple edges.

Input

Input is given from Standard Input in the following format:


N
M_1
a_1 b_1
a_2 b_2
\vdots
a_{M_1} b_{M_1}
M_2
c_1 d_1
c_2 d_2
\vdots
c_{M_2} d_{M_2}
M_3
e_1 f_1
e_2 f_2
\vdots
e_{M_3} f_{M_3}


Output

Print the maximum possible total weight of an independent set in W, modulo 998,244,353.

Examples

Input

2
1
1 2
1
1 2
1
1 2


Output

46494701


Input

3
3
1 3
1 2
3 2
2
2 1
2 3
1
2 1


Output

883188316


Input

100000
1
1 2
1
99999 100000
1
1 100000


Output

318525248

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while (ch<'0'||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do { (t*=10)+=ch-'0'; ch=getchar(); } while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const ll mod=998244353;
const int maxn=(1e5)+10;
int n,mi[maxn],ans;
void update(int &x,int y) {
	x+=y; if (x>=mod) x-=mod;
}
struct Graph {
	int tot,m,head[maxn],nxt[maxn*2],to[maxn*2];
	int sg[maxn],vis[maxn],cnt[maxn];
	void add(int x,int y) {
		tot++; nxt[tot]=head[x];
		head[x]=tot; to[tot]=y;
	}
	void init() {
		read(m);
		int x,y;
		while (m--) {
			read(x); read(y);
			if (x>y) swap(x,y);
			add(x,y);
		}
		tot=0;
		for (int i=n;i>=1;i--) {
			tot++;
			for (int j=head[i];j;j=nxt[j]) vis[sg[to[j]]]=tot;
			for (;vis[sg[i]]==tot;sg[i]++);
			update(cnt[sg[i]],mi[i]);
		}
	}
} T[3];
int main() {
	//freopen("1.txt","r",stdin);
	mi[0]=1; mi[1]=(ll)(1e18)%mod;
	read(n);
	for (int i=2;i<=n;i++) mi[i]=(ll)mi[i-1]*mi[1]%mod;
	for (int i=0;i<3;i++) T[i].init();
	for (int i=0;i<=1000;i++)
	for (int j=0;j<=1000;j++)
		update(ans,(ll)T[0].cnt[i]*T[1].cnt[j]%mod*T[2].cnt[i^j]%mod);
	printf("%d\n",ans);
	return 0;
}
/*
  0. Enough array size? Enough array size? Enough array size? Interger overflow?
  
  1. Think TWICE, Code ONCE!
  Are there any counterexamples to your algo?
    
  2. Be careful about the BOUNDARIES!
  N=1? P=1? Something about 0?
    
  3. Do not make STUPID MISTAKES!
  Time complexity? Memory usage? Precision error?
*/