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 ll long long
using namespace std;
const ll p=998244353;
vector<ll>zh[100010];
ll n,m,i,j,k,x,y,ans,hh,f[3][100010],fac[500010],sg[3][100010];
int main(){
	scanf("%lld",&n);
	fac[0]=1;
	hh=1000000000000000000%p;
	for(i=1;i<=n*3;i++)fac[i]=fac[i-1]*hh%p;
	for(k=0;k<3;k++){
		scanf("%lld",&m);
		for(i=1;i<=n;i++)zh[i].clear();
		for(i=1;i<=m;i++){
			scanf("%lld%lld",&x,&y);
			if(x>y)swap(x,y);
			zh[x].push_back(y);
		}
		for(i=n;i;i--){
			std::set<ll>mex;
			for(j=0;j<zh[i].size();j++)mex.insert(sg[k][zh[i][j]]);
			while(mex.count(sg[k][i]))sg[k][i]++;
			(f[k][sg[k][i]]+=fac[i])%=p;
		}
	}
	for(i=0;i<500;i++)
	 for(j=0;j<500;j++)ans=(ans+f[0][i]*f[1][j]%p*f[2][i^j])%p;
	printf("%lld",ans);
}
/*
观察转移公式，发现跟博弈游戏一样。。
orz 
*/