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 rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using graph=vector<vector<int>>;

class mint{
	static const int MOD=998244353;
	int x;
public:
	mint():x(0){}
	mint(long long y){ x=y%MOD; if(x<0) x+=MOD; }

	mint& operator+=(const mint& m){ x+=m.x; if(x>=MOD) x-=MOD; return *this; }
	mint& operator-=(const mint& m){ x-=m.x; if(x<   0) x+=MOD; return *this; }
	mint& operator*=(const mint& m){ x=1LL*x*m.x%MOD; return *this; }
	mint& operator/=(const mint& m){ return *this*=inverse(m); }
	mint operator+(const mint& m)const{ return mint(*this)+=m; }
	mint operator-(const mint& m)const{ return mint(*this)-=m; }
	mint operator*(const mint& m)const{ return mint(*this)*=m; }
	mint operator/(const mint& m)const{ return mint(*this)/=m; }
	mint operator-()const{ return mint(-x); }

	friend mint inverse(const mint& m){
		int a=m.x,b=MOD,u=1,v=0;
		while(b>0){ int t=a/b; a-=t*b; swap(a,b); u-=t*v; swap(u,v); }
		return u;
	}

	friend istream& operator>>(istream& is,mint& m){ long long t; is>>t; m=mint(t); return is; }
	friend ostream& operator<<(ostream& os,const mint& m){ return os<<m.x; }
	int to_int()const{ return x; }
};

mint operator+(long long y,const mint& m){ return m+y; }
mint operator*(long long y,const mint& m){ return m*y; }

mint pow(mint m,long long k){
	mint res=1;
	for(;k>0;k>>=1,m*=m) if(k&1) res*=m;
	return res;
}

int main(){
	int n; scanf("%d",&n);
	vector<graph> G(3,graph(n));
	rep(i,3){
		int m; scanf("%d",&m);
		rep(j,m){
			int u,v; scanf("%d%d",&u,&v); u--; v--;
			if(u>v) swap(u,v);
			G[i][u].emplace_back(v);
		}
	}

	vector<vector<int>> grundy(3,vector<int>(n));
	rep(i,3){
		vector<int> used(1e5+1,-1);
		for(int u=n-1;u>=0;u--){
			for(int v:G[i][u]){
				used[grundy[i][v]]=u;
				while(used[grundy[i][u]]==u) grundy[i][u]++;
			}
		}
	}

	int g_max=0;
	rep(i,3) rep(u,n) g_max=max(g_max,grundy[i][u]);

	vector<vector<mint>> wt(3,vector<mint>(g_max+1));
	rep(i,3) rep(u,n) wt[i][grundy[i][u]]+=pow(mint(10),18*(u+1));

	mint ans=0;
	rep(g0,g_max+1) rep(g1,g_max+1) {
		int g2=g0^g1;
		if(g2>g_max) continue;
		ans+=wt[0][g0]*wt[1][g1]*wt[2][g2];
	}
	cout<<ans<<'\n';

	return 0;
}
