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 pb push_back
using namespace std;
typedef long long ll;
const int maxn=1e5+3;
const int M=998244353;
const int inv2=(M+1)/2;
const int P=1000000000000000000ll%M;
const int E=1<<17;
vector <int> h[maxn];
int sg[maxn],n,m,cnt[3][E],a[maxn],ans[maxn];
bool f[maxn];
void add(int &x,int y){
    x+=y;if (x>=M)x-=M;
}
void FWT(int a[],int n){
    for(int d=1;d<n;d<<=1)
        for(int m=d<<1,i=0;i<n;i+=m)
            for(int j=0;j<d;j++){
                int x=a[i+j],y=a[i+j+d];
                a[i+j]=(x+y)%M,a[i+j+d]=(x-y+M)%M;
            }
}
void UFWT(int a[],int n){
    for(int d=1;d<n;d<<=1)
        for(int m=d<<1,i=0;i<n;i+=m)
            for(int j=0;j<d;j++){
                int x=a[i+j],y=a[i+j+d];
                a[i+j]=1ll*(x+y)*inv2%M,a[i+j+d]=1ll*(x-y+M)*inv2%M;
            }
}
int main(){
    scanf("%d",&n);
    a[0]=1;
    for (int i=1;i<=n;i++) a[i]=1ll*a[i-1]*P%M;
    for (int i=0;i<3;i++){
        scanf("%d",&m);
        for (int j=1;j<=n;j++) h[j].clear();
        for (int j=0;j<m;j++){
            int u,v;scanf("%d%d",&u,&v);
            if (u>v) swap(u,v);
            h[u].pb(v);
        }
        for (int j=n;j;j--){
            sg[j]=0;
            for (auto v:h[j]) f[sg[v]]=1;
            while (f[sg[j]]) sg[j]++;
            for (auto v:h[j]) f[sg[v]]=0;
            add(cnt[i][sg[j]],a[j]);
        }
        FWT(cnt[i],E);
    }
    for (int i=0;i<E;i++) ans[i]=1ll*cnt[0][i]*cnt[1][i]%M*cnt[2][i]%M;
    UFWT(ans,E);
    printf("%d\n",ans[0]);
    return 0;
}
