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;
const int B=716070898,mod=998244353,N=5e5+50;
int n,m,pw[N],sum[3][N],mx,f[N],ans;bool vis[N];
vector<int>v[N];
void init(int id){
    scanf("%d",&m);
    for(int i=1;i<=n;i++)v[i].clear();
    for(int i=1,x,y;i<=m;i++){
        scanf("%d%d",&x,&y);
        if(x>y)swap(x,y);v[x].pb(y);
    }
    for(int i=n;i;i--){
        for(int j=0;j<v[i].size();j++)vis[f[v[i][j]]]=1;
        for(int j=0;;j++)if(!vis[j]){f[i]=j;break;}
        for(int j=0;j<v[i].size();j++)vis[f[v[i][j]]]=0;
        (sum[id][f[i]]+=pw[i])%=mod;mx=max(mx,f[i]);
    }
}
int main(){
    scanf("%d",&n);pw[0]=1;
    for(int i=1;i<=n;i++)pw[i]=1ll*pw[i-1]*B%mod;
    init(0);init(1);init(2);
    for(int i=0;i<=mx;i++)for(int j=0;j<=mx;j++)
        ans=(ans+1ll*sum[0][i]*sum[1][j]%mod*sum[2][i^j])%mod;
    cout<<ans;
    return 0;
}