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>
#define ll long long
#define uint unsigned int
#define ull unsigned long long
#define inf 1010000000
#define infll 1010000000000000000ll
#define vi vector<int>
#define vll vector<ll>
#define pb push_back
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd pair<double,double>
#define mpr make_pair
#define fi first
#define se second
#define pq priority_queue<int>
#define pqll priority_queue<ll>
#define up(x,y) (x<(y)?x=(y):0)
#define dn(x,y) (x>(y)?x=(y):0)
#define ad(x,y) (x=(x+(y))%mod)
#define cbn(x,y) ((ll)fac[x]*inv[y]%mod*inv[(x)-(y)]%mod)
#define N 100009
#define mod 998244353
using namespace std;

int read(); ll readll();

int n,m,mx,pw[N],a[N],b[N],c[N],f[N],g[N],h[N],tg[N]; vi e[N];
void work(int *a,int *b){
	m=read();
	int i,x,y;
	for (i=1; i<=n; i++) e[i].clear();
	for (i=1; i<=m; i++){
		x=read(); y=read();
		if (x>y) swap(x,y); e[x].pb(y);	
	}
	memset(tg,0,sizeof(tg));
	for (i=n; i; i--){
		for (int p:e[i]) tg[a[p]]=i;
		for (; tg[a[i]]==i; a[i]++);
		up(mx,a[i]); ad(b[a[i]],pw[i]);
	}
}
int main(){
	n=read();
	int i,j;
	for (i=pw[0]=1; i<=n; i++) 
		for (j=0,pw[i]=pw[i-1]; j<18; j++) pw[i]=10ll*pw[i]%mod;
	work(a,f); work(b,g); work(c,h);
	int ans=0;
	for (i=0; i<=mx; i++)
		for (j=0; j<=mx; j++) ad(ans,(ll)f[i]*g[j]%mod*h[i^j]);
	printf("%d\n",ans);
	return 0;
}
int read(){
	int x=0; char ch=getchar(); bool flag=0;
	while (ch<'0' || ch>'9'){ if (ch=='-') flag=1; ch=getchar(); }
	while (ch>='0' && ch<='9'){ x=x*10+ch-'0'; ch=getchar(); }
	return flag?-x:x;
}
ll readll(){
	ll x=0; char ch=getchar(); bool flag=0;
	while (ch<'0' || ch>'9'){ if (ch=='-') flag=1; ch=getchar(); }
	while (ch>='0' && ch<='9'){ x=x*10+ch-'0'; ch=getchar(); }
	return flag?-x:x;
}