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>
using namespace std;
const int MOD = 998244353;

int sum(int a, int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int sub(int a, int b) {
    return a - b < 0 ? a - b + MOD : a - b;
}
int mul(int a, int b) {
    return (1LL*a*b)%MOD;
}
int pw(int a, int b) {
	if (!b) return 1;
	int r = pw(a, b/2);
	r = mul(r, r);
	if (b%2) r = mul(r, a);
	return r;
}

const int MAXN = 3e5+7;
vector< int >G[3][MAXN];
int grundy[3][MAXN];
int fr[3][MAXN];
int mx[3];

int p10[MAXN];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m;
    cin >> n;

    p10[0] = 1;
    const int base = pw(10, 18);
    for (int i = 1; i <= n; i++) p10[i] = mul(base, p10[i-1]);

    for (int c = 0; c < 3; c++) {
        cin >> m;
        while (m--) {
            int u, v;
            cin >> u >> v;
            if (u > v) swap(u, v);
            G[c][u].push_back(v);
        }

        for (int i = n; i > 0; i--) {
            set< int >st;
            for (int v : G[c][i]) st.insert(grundy[c][v]);

            grundy[c][i] = 0;
            for (int x : st) {
                if (x==grundy[c][i]) grundy[c][i]++;
                else break;
            }

//            cout << c << " " << i << " => " << grundy[c][i] << endl;
            fr[c][grundy[c][i]] = sum(fr[c][grundy[c][i]], p10[i]);
            mx[c] = max(mx[c], grundy[c][i]);
        }
    }

    int ans = 0;
    for (int g0 = 0; g0 <= mx[0]; g0++) {
        for (int g1 = 0; g1 <= mx[1]; g1++) {
            int g2 = g0^g1;
            if (g2 > mx[2]) continue;

            int tmp = mul(mul(fr[0][g0], fr[1][g1]), fr[2][g2]);
            ans = sum(ans, tmp);
        }
    }

//    ans = mul(ans, pw(10, 54));

    cout << ans << endl;

    return 0;
}