Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define x first
#define y second
#define mp make_pair
#define pb push_back
template <typename TYPE> inline void chkmax(TYPE &x,TYPE y){x<y?x=y:TYPE();}
template <typename TYPE> inline void chkmin(TYPE &x,TYPE y){y<x?x=y:TYPE();}
template <typename TYPE> void readint(TYPE &x)
{
    x=0;int f=1;char c;
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-1;
    for(;isdigit(c);c=getchar())x=x*10+c-'0';
    x*=f;
}
const int MAXN=100005,MOD=998244353,BASE=716070898;

int n;
vector<int> G[3][MAXN];
int sg[3][MAXN];
int m,a[3][MAXN];
int b[MAXN],c[MAXN],pw[MAXN];

int main()
{
	readint(n);
	pw[0]=1;
	for(int i=1;i<=n;++i)pw[i]=1ll*pw[i-1]*BASE%MOD;
	for(int t=0;t<3;++t)
	{
		int E,u,v;
		readint(E);
		while(E--)readint(u),readint(v),G[t][u].pb(v),G[t][v].pb(u);
		for(int i=n;i;--i)
		{
			unordered_set<int> st;
			for(auto j:G[t][i])if(j>i)st.insert(sg[t][j]);
			for(int k=0;;++k)if(!st.count(k)){sg[t][i]=k;break;}
			a[t][sg[t][i]]=(a[t][sg[t][i]]+pw[i])%MOD;
			chkmax(m,sg[t][i]+1);
		}
	}
	int M;
	for(M=1;M<=m;M<<=1);
	for(int i=0;i<M;++i)
		for(int j=0;j<M;++j)
			c[i^j]=(c[i^j]+1ll*a[0][i]*a[1][j])%MOD;
	for(int i=0;i<M;++i)
		for(int j=0;j<M;++j)
			b[i^j]=(b[i^j]+1ll*c[i]*a[2][j])%MOD;
	printf("%d\n",b[0]);
	return 0;
}