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
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t,class ...A> inline void read(t &x,A &...a){
	read(x);read(a...);
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

const int mod=998244353,N=1e5+5;
int n,cnt[3][600],pw[N*3],ans,sg[N],v[N],tag,m,big;
vector<int> g[N];

int mex(vector<int> &a){
	tag++;
	for(int x:a) v[x]=tag;
	for(int i=0;;i++) if(v[i]^tag) return i;
}

void dfs(int x){
	vector<int> tp;
	for(int y:g[x]){
		if(sg[y]==-1) dfs(y);
		tp.push_back(sg[y]);
	}
	sg[x]=mex(tp);
}

void init(int T){
	read(m);
	for(int i=1;i<=n;i++) g[i].clear(),sg[i]=-1;
	for(int i=1,x,y;i<=m;i++){
		read(x,y);
		if(x>y) swap(x,y);
		g[x].emplace_back(y);
	}
	for(int i=1;i<=n;i++) if(sg[i]==-1) dfs(i);
	for(int i=1;i<=n;i++) cnt[T][sg[i]]=(cnt[T][sg[i]]+pw[i])%mod;
}

signed main(){
	read(n);
	pw[0]=big=1;
	for(int i=1;i<=18;i++) big=10ll*big%mod;
	for(int i=1;i<=3*n;i++) pw[i]=1ll*pw[i-1]*big%mod;
	for(int i:{0,1,2}) init(i);
	for(int i=0;i<512;i++) for(int j=0;j<512;j++)
		ans=(ans+1ll*cnt[0][i]*cnt[1][j]%mod*cnt[2][i^j])%mod;
	write(ans);
}