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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int MAXN = (1 << 17);
const int MOD = 998244353;
const int INV = (MOD + 1) >> 1;
const int POW = (long long)(1E18) % MOD;

void fwt(int *a, int n, int type) {
    for(int s=2;s<=n;s<<=1) {
        int t = (s>>1);
        for(int i=0;i<n;i+=s) {
            for(int j=0;j<t;j++) {
                int x = a[i+j], y = a[i+j+t];
                a[i+j] = (type == -1) ? 1LL*(x + y)%MOD*INV%MOD : (x + y)%MOD;
                a[i+j+t] = (type == -1) ? 1LL*(x + MOD - y)%MOD*INV%MOD : (x + MOD - y)%MOD;
            }
        }
    }
}

vector<int>G[MAXN + 5];
int mex[MAXN + 5], sg[MAXN + 5];

int ans[MAXN + 5], tmp[MAXN + 5], pw[MAXN + 5];
int main() {
	pw[0] = 1;
	for(int i=1;i<=MAXN;i++)
		pw[i] = 1LL*pw[i-1]*POW%MOD;
		
	int N; scanf("%d", &N);
	int len; for(len = 1; len < N; len <<= 1);
	ans[0] = 1; fwt(ans, len, 1);
	for(int o=0;o<3;o++) {
		int M; scanf("%d", &M);
		for(int i=1;i<=N;i++) G[i].clear();
		for(int i=1;i<=M;i++) {
			int u, v; scanf("%d%d", &u, &v);
			if( u > v ) swap(u, v);
			G[u].push_back(v);
		}
		for(int i=N;i>=1;i--) {
			for(int j=0;j<G[i].size();j++)
				mex[sg[G[i][j]]] = true;
			sg[i] = 0;
			while( mex[sg[i]] ) sg[i]++;
			for(int j=0;j<G[i].size();j++)
				mex[sg[G[i][j]]] = false;
		}
		for(int i=0;i<len;i++) tmp[i] = 0;
		for(int i=1;i<=N;i++) tmp[sg[i]] = (tmp[sg[i]] + pw[i]) % MOD;
		fwt(tmp, len, 1);
		for(int i=0;i<len;i++) ans[i] = 1LL*ans[i]*tmp[i]%MOD;
	}
	fwt(ans, len, -1), printf("%d\n", ans[0]);
}