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;
#define LL long long
const int MAXN = 1e5 + 11;
const int MAXM = 555;
const int mod = 998244353;
#define FOR(i, a, b) for(int i = (a);i <= (b); i++)
#define ROF(i, a, b) for(int i = (a);i >= (b); i--)
int n, m;
int f[3][MAXM], sg[MAXN], mark[MAXM];
vector<int> G[MAXN];
void Add(int &x, int y){
    x = x + y - (x + y >= mod ? mod : 0);
}
void Mu(int &x, int y){
    x = 1LL * x * y % mod;
}
int A(int x, int y){
    return x + y - (x + y >= mod ? mod : 0);
}
int M(int x, int y){
    return 1LL * x * y % mod;
}
int ksm(int base, int x){
    base %= mod;
    int ans = 1;
    while(x){
        if(x & 1)Mu(ans, base);
        x >>= 1;
        Mu(base, base);
    }
    return ans;
}
int main(){
    cin>>n;
    FOR(t, 0, 2){
        cin>>m;
        int u, v;
        memset(sg, -1, sizeof sg);
        memset(mark, 0, sizeof mark);
        FOR(i, 1, m){
            scanf("%d%d", &u, &v);
            if(u < v)swap(u, v);
            G[v].push_back(u);
        }
        ROF(u, n, 1){
            for(int v : G[u]){
                mark[sg[v]] = u;
            }
            sg[u] = 0;
            while(mark[sg[u]] == u)sg[u]++;
            Add(f[t][sg[u]], ksm((LL)(1e18 + 0) % mod, u));
            G[u].clear();
            //printf("u=%d sg=%d\n", u, sg[u]);
        }
    }
    int ans = 0;
    FOR(i, 0, 500){
        FOR(j, 0, 500){
            Add(ans, M(M(f[0][i],f[1][j]), f[2][i^j]));
        }
    }
    cout<<ans<<endl;
    return 0;
}
