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>
#define ll long long
using namespace std;
const int MAXN=100010;
const ll P=998244353,T=1000000000000000000%P;
int n,x,y,eg[3],m[3],hd[3][MAXN],nx[3][MAXN*2],ver[3][MAXN*2],mx[3],sg[3][MAXN],tmp[MAXN];
ll cnt[3][MAXN],ans;
ll qpow (ll a,ll b) {
	ll res=1;
	while (b) {
		if (b&1) {res=(res*a)%P;}
		a=(a*a)%P,b>>=1;
	}
	return res;
}
void add_edge (int t,int x,int y) {
	ver[t][++eg[t]]=y;
	nx[t][eg[t]]=hd[t][x];
	hd[t][x]=eg[t];
	return;
}
int main () {
	scanf("%d",&n);
	for (int i=0;i<=2;i++) {
		scanf("%d",&m[i]);
		for (int j=1;j<=m[i];j++) {
			scanf("%d%d",&x,&y);
			add_edge(i,min(x,y),max(x,y));
		}
		for (int j=n;j>=1;j--) {
			int cur=0;
			for (int k=hd[i][j];k;k=nx[i][k]) {
				tmp[sg[i][ver[i][k]]]++;
				while (tmp[cur]) {cur++;}
			}
			sg[i][j]=cur;
			for (int k=hd[i][j];k;k=nx[i][k]) {tmp[sg[i][ver[i][k]]]--;}
			cnt[i][cur]=(cnt[i][cur]+qpow(T,j))%P,mx[i]=max(mx[i],cur);
		}
	}
	for (int i=0;i<=mx[0];i++) {
		for (int j=0;j<=mx[1];j++) {
			ans=(ans+((cnt[0][i]*cnt[1][j])%P)*cnt[2][i^j])%P;
		}
	}
	printf("%lld\n",ans);
	return 0;
}