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 N=1e5+5,M=505,mod=998244353;
ll base1=1,base2=0,p[N];
int n,m,tot,vis[N],head[N],nex[N<<1],to[N<<1];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
vector<ll>res[3];
void solve(int flag)
{
    tot=0;memset(head,0,sizeof(head));
    memset(vis,false,sizeof(vis));
    m=100000;
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        //u=v=1;while(u==v) u=rand()*rand()%n+1,v=rand()*rand()%n+1;
        add(u,v);add(v,u);
    }
    priority_queue<int>q[2];
    int x=0,tim=0;
    for(int i=1;i<=n;i++) q[x].push(i);
    vector<ll>f;
    while(!q[x].empty())
    {
        tim++;
        ll ans=0;
        while(!q[x].empty())
        {
            while(!q[x].empty()&&vis[q[x].top()]==tim) q[x].pop();
            if(q[x].empty()) break;
            int u=q[x].top();q[x].pop();
            vis[u]=1e9;
            ans=(ans+p[u])%mod;
            for(int i=head[u];i;i=nex[i])
            {
                int v=to[i];
                if(vis[v]<tim)
                {
                    vis[v]=tim;
                    q[x^1].push(v);
                }
            }
        }
        x^=1;
        f.push_back(ans);
    }
    res[flag]=f;
}
bool vis1[M][M],vis2[M];
ll solve()
{
    memset(vis2,0,sizeof(vis2));
    ll ans=0;
    for(int i=0;i<res[1].size();i++)
    {
        int mx=0;
        while(mx<res[2].size()&&vis1[i][mx]||vis2[mx]) mx++;
        if(mx<res[2].size())
        {
            vis1[i][mx]=vis2[mx]=true;
            ans=(ans+res[1][i]*res[2][mx])%mod;
        }
    }
    return ans;
}
int main()
{
    srand(time(0));
    p[0]=1;
    for(int i=1;i<N;i++) p[i]=1000000000000000000ll%mod*p[i-1]%mod;
    scanf("%d",&n);
    solve(0);solve(1);solve(2);
    assert(res[0].size()<M);
    assert(res[1].size()<M);
    assert(res[2].size()<M);
    ll ans=0;
    for(int i=0;i<res[0].size();i++)
        ans=(ans+res[0][i]*solve())%mod;
    printf("%lld\n",ans);
}
