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;
const int mod=998244353;
int n,m,x,y,sg1[111111],sg2[111111],sg3[111111],pos,ans;
long long a[111111],b[111111],c[111111];
vector<int> g1[111111],g2[111111],g3[111111];
bool used[111111];
int binpow(int a,int t)
{
	int res=1,p=a;
	for (int i=t;i;i>>=1)
	{
		if (i&1) res=1ll*res*p%mod;
		p=1ll*p*p%mod;
	}
	return res;
}
int main()
{
	scanf("%d",&n);
	scanf("%d",&m);
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		if (x>y) swap(x,y);
		g1[x].push_back(y);
	}
	for (int i=n;i>=1;i--)
	{
		for (int j=0;j<g1[i].size();j++)
		{
			used[sg1[g1[i][j]]]=1;
		}
		pos=0;
		while(used[pos]) pos++;
		sg1[i]=pos;a[pos]=(a[pos]+binpow(10,18*i))%mod;
		for (int j=0;j<g1[i].size();j++)
		{
			used[sg1[g1[i][j]]]=0;
		}
	}
	scanf("%d",&m);
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		if (x>y) swap(x,y);
		g2[x].push_back(y);
	}
	for (int i=n;i>=1;i--)
	{
		for (int j=0;j<g2[i].size();j++)
		{
			used[sg2[g2[i][j]]]=1;
		}
		pos=0;
		while(used[pos]) pos++;
		sg2[i]=pos;b[pos]=(b[pos]+binpow(10,18*i))%mod;
		for (int j=0;j<g2[i].size();j++)
		{
			used[sg2[g2[i][j]]]=0;
		}
	}
	scanf("%d",&m);
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		if (x>y) swap(x,y);
		g3[x].push_back(y);
	}
	for (int i=n;i>=1;i--)
	{
		for (int j=0;j<g3[i].size();j++)
		{
			used[sg3[g3[i][j]]]=1;
		}
		pos=0;
		while(used[pos]) pos++;
		sg3[i]=pos;c[pos]=(c[pos]+binpow(10,18*i))%mod;
		for (int j=0;j<g3[i].size();j++)
		{
			used[sg3[g3[i][j]]]=0;
		}
	}
	for (int i=0;i<=n;i++)
	{
		if (!a[i]) continue;
		for (int j=0;j<=n;j++)
		{
			if (!b[j] || !c[i^j]) continue;
			ans=(1ll*a[i]*b[j]%mod*c[i^j]+ans)%mod;
		}
	}
	printf("%d\n",ans);
	return 0;
}