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
// C.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 1e5 + 200, mod = 998244353;

typedef long long ll;

int n, m[3], base, vis[MAX_N], sg[MAX_N];
vector<int> G[MAX_N], sum[3];

int fpow(int bas, int tim)
{
    int ret = 1;
    while (tim)
    {
        if (tim & 1)
            ret = 1LL * ret * bas % mod;
        bas = 1LL * bas * bas % mod;
        tim >>= 1;
    }
    return ret;
}

int main()
{
    scanf("%d", &n), base = fpow(ll(1e18) % mod, mod - 2);
    for (int rd = 0; rd < 3; rd++)
    {
        scanf("%d", &m[rd]);
        for (int i = 1; i <= n; i++)
            G[i].clear();
        for (int i = 1, u, v; i <= m[rd]; i++)
        {
            scanf("%d%d", &u, &v);
            if (u > v)
                swap(u, v);
            G[u].push_back(v);
        }
        memset(sg, 0, sizeof(sg)), memset(vis, -1, sizeof(vis));
        int cbase = fpow(10, 18 * n);
        for (int i = n; i >= 1; i--)
        {
            for (int v : G[i])
                vis[sg[v]] = i;
            while (vis[sg[i]] == i)
                sg[i]++;
            while (sum[rd].size() <= sg[i])
                sum[rd].push_back(0);
            sum[rd][sg[i]] = (0LL + sum[rd][sg[i]] + cbase) % mod, cbase = 1LL * cbase * base % mod;
        }
    }
    int ans = 0;
    for (int i = 0, siz1 = sum[0].size(); i < siz1; i++)
        for (int j = 0, siz2 = sum[1].size(); j < siz2; j++)
        {
            int k = i ^ j;
            if (k < sum[2].size())
                ans = (0LL + ans + 1LL * sum[0][i] * sum[1][j] % mod * sum[2][k] % mod) % mod;
        }
    printf("%d\n", ans);
    return 0;
}