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;
#define ll long long
#define ui unsigned int
#define ull unsigned long long
#define db long double
#define pii pair<int,ll>
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define vi vector<int>
#define lb lower_bound
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,b,a) for(int i=(b);i>=(a);--i)
#define rep0(i,a,b) for(int i=(a);i<(b);++i)
#define fore(i,a) for(int i=0;i<a.size();++i)
#define ls x<<1,l,m
#define rs x<<1|1,m+1,r
#define gc() getchar()
inline int rd()
{
	int x=0,w=1;char c=gc();while(!isdigit(c)&&c!='-')c=gc();
	if(c=='-')w=-1,c=gc();while(isdigit(c))x=x*10+c-48,c=gc();return x*w;
}
const int N=100005,M=1<<19,P=998244353,iv2=(P+1)/2;
int n,s[N],sg[N],a[M],b[M],c[M],d[M];vi g[N];
inline int pw(int a,int b){int r=1;for(;b;b>>=1,a=1ll*a*a%P)if(b&1)r=1ll*r*a%P;return r;}
inline void fwt(int n,int*a,int o)
{
	for(int i=2;i<=n;i<<=1)for(int j=0;j<n;j+=i)rep0(k,0,(i>>1))
	{
		int x=a[j+k],y=a[j+k+(i>>1)];
		a[j+k]=(x+y)%P;a[j+k+(i>>1)]=(x+P-y)%P;
		if(o==-1)
		{
			a[j+k]=1ll*a[j+k]*iv2%P;
			a[j+k+(i>>1)]=1ll*a[j+k+(i>>1)]*iv2%P;
		}
	}
}
void calc(int n,int*a)
{
	int m=rd();rep(i,1,n)g[i].clear();
	rep(i,1,m){int u=rd(),v=rd();if(u>v)swap(u,v);g[u].pb(v);}
	per(i,n,1)
	{
		for(int x:g[i])s[sg[x]]++;
		sg[i]=0;while(s[sg[i]])sg[i]++;
		for(int x:g[i])s[sg[x]]--;
		a[sg[i]]=(a[sg[i]]+pw(10,18*i))%P;
	}
}
int main()
{
	n=rd();calc(n,a);calc(n,b);calc(n,c);
	fwt(M,a,1);fwt(M,b,1);fwt(M,c,1);
	rep0(i,0,M)d[i]=1ll*a[i]*b[i]%P*c[i]%P;
	fwt(M,d,-1);printf("%d\n",d[0]);return 0;
}