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 lint;
typedef long double louble;

#define so(a) ((int)((a).size()))

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

const char lf = '\n';

namespace ae86
{
	const int bufl = 1 << 15;

	char buf[bufl],*s=buf,*t=buf;

	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}

	inline int ty()
	{
		int a=0;int b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}
}
using ae86::ty;

const int _ = 100007 , D = 3 , __ = _ * D , LIM = 512 , mo = 998244353;
const lint linf = 1000000000000000000;

template<typename T1,typename T2> inline T1 ad(T1 &a,T2 b){return a=a+b>=mo?a+b-mo:a+b;}
template<typename T1,typename T2> inline T1 dl(T1 &a,T2 b){return a=a>=b?a-b:a-b+mo;}
template<typename T1,typename T2> inline T1 add(T1 a,T2 b){return a+b>=mo?a+b-mo:a+b;}
template<typename T1,typename T2> inline T1 del(T1 a,T2 b){return a>=b?a-b:a-b+mo;}

lint powa(lint a,lint t)
{
	lint b=1;
	while(t){if(t&1)b=b*a%mo;a=a*a%mo,t>>=1;}
	return b;
}

inline lint inva(lint a){return powa(a,mo-2);}

vector<int> e[_];
lint bas[__];
int n,sg[_];

void makef(lint f[])
{
	int m=ty();
	for(int i=1;i<=n;i++)sg[i]=0,e[i].clear();
	for(int i=1,a,b;i<=m;i++)a=ty(),b=ty(),e[min(a,b)].emplace_back(max(a,b));
	for(int i=n;i>=1;i--)
	{
		unordered_set<int> hav;
		for(auto b:e[i])hav.emplace(sg[b]);
		sg[i]=0;
		while(hav.count(sg[i]))sg[i]++;
	}
	for(int i=0;i<LIM;i++)f[i]=0;
	for(int i=1;i<=n;i++)ad(f[sg[i]],bas[i]);	
}

lint f[D][LIM];

int main()
{
	ios::sync_with_stdio(0),cout.tie(nullptr);

	n=ty();
	bas[0]=1;
	for(int i=1;i<__;i++)bas[i]=bas[i-1]*(linf%mo)%mo;
	for(int i=0;i<D;i++)makef(f[i]);

	lint ans=0;
	for(int i=0;i<LIM;i++)
		for(int j=0;j<LIM;j++)
			ad(ans,f[0][i]*f[1][j]%mo*f[2][i^j]%mo);
	cout<<ans<<lf;
	
	return 0;
}