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 MAXN 100005
#define MAXM 525
#define INF 1000000000000000000
#define MOD 998244353
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n,m[3];
vector<int> G[MAXN];
int sg[3][MAXN];
int weight[3][MAXM];
bool has[MAXM];
int p[3*MAXN];
void add(int &a,int b) {a+=b; if(a>=MOD) a-=MOD;}
void dec(int &a,int b) {a-=b; if(a<MOD) a+=MOD;}
int main()
{
    ll x=INF%MOD;
    p[0]=1;
    for(int i=1;i<=300000;i++) p[i]=1LL*p[i-1]*x%MOD;
    scanf("%d",&n);
    for(int i=0;i<3;i++)
    {
        scanf("%d",&m[i]);
        for(int j=1;j<=n;j++) G[j].clear();
        for(int j=0;j<m[i];j++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(x>y) swap(x,y);
            G[x].push_back(y);
        }
        for(int j=n;j>=1;j--)
        {
            memset(has,false,sizeof(has));
            for(auto to:G[j]) has[sg[i][to]]=true;
            int k=0;
            for(;has[k];k++);
            sg[i][j]=k;
            add(weight[i][k],p[j]);
        }
    }
    int ans=0;
    for(int i=0;i<512;i++)
        for(int j=0;j<512;j++)
        {
            int k=i^j;
            add(ans,1LL*weight[0][i]*weight[1][j]%MOD*weight[2][k]%MOD);
        }
    printf("%d\n",ans);
    return 0;
}