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 <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
const int N=1e6+10;
const int mod=998244353;
int h[N],e[N],ne[N],idx;
int f[N],g[N];
ll sum[3][N];
int sz[3];
void add(int a,int b)
{
    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
int inv;
ll qmi(int a,int b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)ans=(ans*a)%mod;
        a=((ll)a*a)%mod;
        b>>=1;
    }
    return ans;
}
ll gs(int a,int b)
{
    return qmi(a,b);
}
int main()
{
    int n;
    cin>>n;
    inv=qmi(qmi(10,18),mod-2);
    for(int i=0;i<3;i++)
    {
        int m;
        cin>>m;
        memset(h,-1,sizeof h);
        idx=0;
        for(int j=0;j<m;j++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            if(a>b)swap(a,b);
            add(a,b);
        }
        memset(g,0,sizeof g);
        ll now=gs(10,18*n);
        int num=0;
        for(int j=n;j>=1;j--)
        {
            num++;
            for(int k=h[j];k!=-1;k=ne[k])
            {
                g[f[e[k]]]=num;
            }
            f[j]=0;
            while(g[f[j]]==num)f[j]++;
            // cout<<f[j]<<endl;
            sz[i]=max(sz[i],f[j]);
            ll&v=sum[i][f[j]];
            // cout<<now<<endl;
            v=(v+now)%mod;
            now=(now*inv)%mod;
        }
    }
    ll ans=0;
    // cout<<sz[1]<<endl;
    for(int i=0;i<=sz[0];i++)
    {
        for(int j=0;j<=sz[1];j++)
        {
            int k=i^j;
            ans=(ans+sum[0][i]*sum[1][j]%mod*sum[2][k]%mod)%mod;
        }
    }
    cout<<ans<<endl;
}