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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(a) int(a.size())
const int N=1e5+10;
const int M=500;
const int mod=998244353;

int gi() {
	int x=0,o=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*o;
}

void inc(int &a,int b) {
	a=a+b>=mod?a+b-mod:a+b;
}

int pw[N],n;

struct graph {
	vi E[N];
	int vis[N],sg[N];

	void adde(int u,int v) {
		E[u].pb(v);E[v].pb(u);
	}

	vi cal() {
		int m=gi();
		vi ret;
		for(int i=1,u,v;i<=m;i++) u=gi(),v=gi(),adde(u,v);
		for(int u=n;u;u--) {
			for(auto v:E[u]) if(u<v) vis[sg[v]]=u;
			for(int i=0;;i++)
				if(vis[i]!=u) { sg[u]=i;break; }
			while(sz(ret)<=sg[u]) ret.pb(0);
			inc(ret[sg[u]],pw[u]);
		}
		return ret;
	}
} G1,G2,G3;

int main() {
	pw[0]=1;
	pw[1]=1;for(int i=1;i<=18;i++) pw[1]=10ll*pw[1]%mod;
	for(int i=2;i<N;i++) pw[i]=1ll*pw[i-1]*pw[1]%mod;
	cin>>n;
	vi v1=G1.cal(),v2=G2.cal(),v3=G3.cal();
	int ans=0;
	for(int i=0;i<sz(v1);i++)
		for(int j=0;j<sz(v2);j++) {
			int k=i^j;
			if(k<sz(v3)) inc(ans,1ll*v1[i]*v2[j]%mod*v3[k]%mod);
		}
	cout<<ans;
	return 0;
}
