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>
using namespace std;
typedef long long ll;
const int max_n=100005;
const int mod=998244353;
namespace {
    inline int Add(int x,int y){return (x+=y)>=mod?x-mod:x;}
    inline int Sub(int x,int y){return (x-=y)<0?x+mod:x;}
    inline int Mul(int x,int y) {return 1ll*x*y%mod;}
    inline int Pow(int x,int y=mod-2){if(y==0)return 1;int res=1;while(y){if(y&1)res=1ll*res*x%mod;x=1ll*x*x%mod;y>>=1;}return res;}
}
int n,m;
vector<int> G[max_n];
vector<int> solve(vector<int> a,vector<int> b){
    vector<int> c;
    c.assign(1024,0);
    for(int i=0;i<1024;i++){
        for(int j=0;j<1024;j++)c[i^j]=Add(c[i^j],Mul(a[i],b[j]));
    }
    return c;
}
int p[max_n];
int sg[max_n];
int SG(int u){
    if(~sg[u])return sg[u];
    set<int> s;
    for(auto v:G[u]){
        if(v>u)s.insert(SG(v));
    }
    for(int i=0;;i++)if(s.count(i)==0)return sg[u]=i;
}
int main(){
    p[0]=1,p[1]=Pow(10,18);
    for(int i=2;i<max_n;i++)p[i]=Mul(p[i-1],p[1]);
    scanf("%d",&n);
    vector<int> f;f.assign(1024,0);f[0]=1;
    vector<int> a;
    scanf("%d",&m);
    for(int i=0;i<m;i++){
        int u,v;scanf("%d%d",&u,&v);
        G[u].push_back(v),G[v].push_back(u);
    }
    for(int i=1;i<=n;i++)sg[i]=-1;
    a.assign(1024,0);
    for(int i=n;i>=1;i--){
        int x=SG(i);
        assert(x<1024);
        a[x]=Add(a[x],p[i]);
    }
    f=solve(a,f);
    scanf("%d",&m);
    for(int i=1;i<=n;i++)G[i].clear();
    for(int i=0;i<m;i++){
        int u,v;scanf("%d%d",&u,&v);
        G[u].push_back(v),G[v].push_back(u);
    }
    for(int i=1;i<=n;i++)sg[i]=-1;
    a.assign(1024,0);
    for(int i=n;i>=1;i--){
        int x=SG(i);
        assert(x<1024);
        a[x]=Add(a[x],p[i]);
    }
    f=solve(a,f);
    scanf("%d",&m);
    for(int i=1;i<=n;i++)G[i].clear();
    for(int i=0;i<m;i++){
        int u,v;scanf("%d%d",&u,&v);
        G[u].push_back(v),G[v].push_back(u);
    }
    for(int i=1;i<=n;i++)sg[i]=-1;
    a.assign(1024,0);
    for(int i=n;i>=1;i--){
        int x=SG(i);
        assert(x<1024);
        a[x]=Add(a[x],p[i]);
    }
    f=solve(a,f);
    printf("%d\n",f[0]);
    return 0;
}