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 pair<int,int> PII;
const int maxn=100010,mod=998244353,w=1000000000000000000%mod;
#define MP make_pair
#define PB push_back
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline ll read(){
	char ch=getchar();ll x=0,f=0;
	while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
	return f?-x:x;
}
inline int qmo(int x){return x+(x>>31)*mod;}
int n,m,sq[3],sg[3][maxn],sum[3][maxn],cnt[maxn],el,head[maxn],to[maxn],nxt[maxn],pw[maxn],ans;
inline void add(int u,int v){
	to[++el]=v;nxt[el]=head[u];head[u]=el;
}
int main(){
	n=read();
	pw[0]=1;
	FOR(i,1,n) pw[i]=1ll*pw[i-1]*w%mod;
	FOR(_,0,2){
		m=read();
		sq[_]=sqrt(2*m);
		FOR(i,1,m){
			int u=read(),v=read();
			if(u>v) swap(u,v);
			add(u,v);
		}
		ROF(i,n,1){
			for(int j=head[i];j;j=nxt[j]) cnt[sg[_][to[j]]]++;
			while(cnt[sg[_][i]]) sg[_][i]++;
			for(int j=head[i];j;j=nxt[j]) cnt[sg[_][to[j]]]--;
		}
		FOR(i,1,n) sum[_][sg[_][i]]=(sum[_][sg[_][i]]+pw[i])%mod;
		FOR(i,1,n) head[i]=0;
		FOR(i,1,el) to[i]=nxt[i]=0;
		el=0;
	}
	FOR(i,0,sq[0]) FOR(j,0,sq[1]) ans=(ans+1ll*sum[0][i]*sum[1][j]%mod*sum[2][i^j])%mod;
	printf("%d\n",ans);
}