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 Fail puts("NO"),exit(0);
#define eps 1e-10
#define maxn 100010
#define maxm 500010
#define inf 1000000007
#define mod 998244353
#define pi acos(-1)
#define mp(x,y) make_pair(x,y)
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int>pii;
typedef bitset<maxn> bit;
int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch-'0'<0||ch-'0'>9){if(ch=='-') f=-1;ch=getchar();}
    while(ch-'0'>=0&&ch-'0'<=9){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;
vector<int>e[3][maxn];
int val[maxn],sg[3][maxn],f[3][maxn];
int main()
{
    n=read();
    val[0]=1;
    int base=(int)(1000000000000000000ll%mod);
    for(int i=1;i<=n;i++)  val[i]=1ll*val[i-1]*base%mod;
    for(int i=0;i<3;i++)
    {
        int m=read();
        while(m--)
        {
            int u=read(),v=read();
            if(u<v)  e[i][u].push_back(v);
            else     e[i][v].push_back(u);
        }
        for(int j=n;j>=1;j--)
        {
            if(e[i][j].size())
            {
                set<int>s;
                for(int k=0;k<e[i][j].size();k++)  s.insert(sg[i][e[i][j][k]]);
                while(s.count(sg[i][j]))  sg[i][j]++;
            }
        }
        for(int j=1;j<=n;j++)  f[i][sg[i][j]]=(f[i][sg[i][j]]+val[j])%mod;
    }
    int ans=0;
    for(int i=0;i<500;i++)
      for(int j=0;j<500;j++)  ans=(ans+1ll*f[0][i]*f[1][j]%mod*f[2][i^j])%mod;
    cout<<ans<<endl;
    return 0;
}