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;
#define N 120000

const LL mod=998244353;
LL n,m,p,f[3][N],F[N],pw[N*18],sg[N];
bool vis[N];
vector<LL> g[N];

void upd(LL &x,LL y){x=(x+y)%mod;}

void solve(LL op){
	scanf("%lld",&m);
	for (LL i=1;i<=n;++i) g[i].clear();
	for (LL i=1;i<=m;++i){
		LL x,y; scanf("%lld%lld",&x,&y);
		if (x>y) swap(x,y);
		g[x].push_back(y);
	}
	for (LL i=n;i;--i){
		for (LL j:g[i]) vis[sg[j]]=1;
		for (sg[i]=0;vis[sg[i]];++sg[i]);
		upd(f[op][sg[i]],pw[i*18]);
		for (LL j:g[i]) vis[sg[j]]=0;
	}
}

int main(){
	scanf("%lld",&n);
	pw[0]=1; for (LL i=1;i<=n*18;++i) pw[i]=pw[i-1]*10%mod;
	solve(0); solve(1); solve(2);
	p=600;
	for (LL i=1;i<=2;++i){
		memset(F,0,sizeof F);
		for (LL j=0;j<=p;++j)
			for (LL k=0;k<=p;++k){
				upd(F[j^k],f[i][j]*f[i-1][k]);
			}
		for (LL j=0;j<=p;++j) f[i][j]=F[j];
	}
	printf("%lld\n",f[2][0]);
	
	return 0;
}
			