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;

const int maxn = 100010, P = 998244353;
int n, ans, cnt1[maxn], cnt2[maxn], cnt3[maxn];
int mx, m, sg[maxn];
vector<int> G[maxn];

int qp(int x, int y) {
    int z = 1;
    for (; y; y >>= 1, x = 1LL * x * x % P) {
        if (y & 1) z = 1LL * z * x % P;
    }
    return z;
}

int main() {
    scanf("%d", &n);
    auto solve = [&](int *cnt) {
        memset(sg, 0, sizeof(sg));
        for (int i = 1; i <= n; i++) {
            G[i].clear();
        }
        scanf("%d", &m);
        for (int i = 1, u, v; i <= m; i++) {
            scanf("%d %d", &u, &v);
            G[min(u, v)].push_back(max(u, v));
        }
        for (int i = n; i; i--) {
            unordered_set<int> S;
            for (int j : G[i]) S.insert(sg[j]);
            while (S.count(sg[i])) sg[i]++;
            mx = max(mx, sg[i]);
            (cnt[sg[i]] += qp(10, 18 * i)) %= P;
        }
    };
    solve(cnt1), solve(cnt2), solve(cnt3);
    for (int i = 0; i <= mx; i++) {
        for (int j = 0; j <= mx; j++) {
            ans = (ans + 1LL * cnt1[i] * cnt2[j] % P * cnt3[i ^ j]) % P;
        }
    }
    printf("%d\n", ans);
    return 0;
}