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 ll long long
#define P 998244353
#define B ((ll)1e18%P)
#define MAX 500005
using namespace std;

int n, m, cnt, id;
int head[MAX], vet[MAX], Next[MAX], sg[3][MAX];
ll p[MAX], f[3][MAX];

void add(int x, int y){
    cnt++;
    Next[cnt] = head[x];
    head[x] = cnt;
    vet[cnt] = y;
}

int dfs(int x){
    if(sg[id][x] != -1) return sg[id][x];
    sg[id][x] = 0;
    map<int, bool> mp;
    for(int i = head[x]; i; i = Next[i]){
        int v = vet[i];
        mp[dfs(v)] = true;
    }
    while(mp[sg[id][x]]) sg[id][x]++;
    return sg[id][x];
}

int main()
{
    cin >> n;
    p[0] = 1;
    for(int i = 1; i < MAX; i++){
        p[i] = p[i-1]*B%P;
    }
    memset(sg, -1, sizeof(sg));
    for(id = 0; id < 3; id++){
        memset(head, 0, sizeof(head));
        cnt = 0;
        cin >> m;
        int x, y;
        for(int i = 1; i <= m; i++){
            scanf("%d%d", &x, &y);
            if(x > y) swap(x, y);
            add(x, y);
        }
        for(int i = 1; i <= n; i++){
            if(sg[id][i] == -1) dfs(i);
        }
        for(int i = 1; i <= n; i++){
            (f[id][sg[id][i]] += p[i]) %= P;
        }
    }

    ll ans = 0;
    for(int i = 0; i < 500; i++){
        for(int j = 0; j < 500; j++){
            ans = (ans + f[0][i]*f[1][j]%P*f[2][i^j]%P)%P;
        }
    }
    cout << ans << endl;

    return 0;
}