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 int long long
using namespace std;
inline int read(){
	register int x=0;
	register bool f=0;
	register char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-') f=1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<3)+(x<<1)+c-48;
		c=getchar();
	}
	return f?-x:x;
}
char cr[200],str[200];int tt;
inline void print(int x,char k='\n') {
    if(!x) putchar('0');
    if(x < 0) putchar('-'),x=-x;
    while(x) cr[++tt]=x%10+'0',x/=10;
    while(tt) putchar(cr[tt--]);
    putchar(k);
}
const int maxn=300000+10;
const int p=998244353;
const int b=1000000000000000000ll%p;
int h[maxn],cnt,s[maxn],n;
int sg[maxn],f[4][maxn],m;
bool vis[maxn];
struct _{
	int nxt,to;
}e[maxn<<1];
void add(int u,int v){
	e[++cnt].nxt=h[u];
	e[cnt].to=v;
	h[u]=cnt;
}
signed main(){
    n=read();
    s[0]=1;
    for(int i=1;i<=n*3;i++){
    	s[i]=s[i-1]*b%p;
	}
	for(int o=1;o<=3;o++){
		memset(h,0,sizeof(h));
		memset(sg,0,sizeof(sg));
		cnt=0;m=read();
		for(int i=1;i<=m;i++){
            int u=read(),v=read();
            if(u<v) add(u,v);
            if(v<u) add(v,u);
        }
        for(int i=n;i>=1;i--){
        	for(int j=h[i];j;j=e[j].nxt){
        		vis[sg[e[j].to]]=1;
			}
			for(int j=0;;j++){
				if(!vis[j]){
					sg[i]=j;
					break;
				}
			}
			for(int j=h[i];j;j=e[j].nxt){
        		vis[sg[e[j].to]]=0;
			}
            (f[o][sg[i]]+=s[i])%=p;
		}
	}
	int ans=0;
	for (int i=0;i<500;i++)
        for (int j=0;j<500;j++)
            (ans+=f[3][i]*f[1][j]%p*f[2][i^j])%=p;
    print(ans);
	return 0;
}