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 <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i)
#define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;})
using namespace std;
typedef long long ll;
const int N = 3e5+10, P = 998244353;
int n, m, vis[N], sg[N], h[N], f[3][N];
vector<int> g[N];
void add(int &a, ll b) {a=(a+b)%P;}

int main() {
	h[0] = 1;
	h[1] = 1000000000000000000ll%P;
	REP(i,2,N-1) h[i] = (ll)h[i-1]*h[1]%P;
	scanf("%d", &n);
int mx = 0;
	REP(i,0,2) {
		scanf("%d", &m);
mx = max(mx, int(sqrt(1+4*m)/2));
		REP(j,1,n) g[j].clear();
		REP(j,1,m) {
			int u,v;
			scanf("%d%d", &u, &v);
			if (u>v) swap(u,v);
			g[u].pb(v);
		}
		PER(j,1,n) {
			int sz = g[j].size();
			REP(k,0,sz) vis[k] = 0;
			for (int k:g[j]) vis[sg[k]] = 1;
			REP(k,0,sz) if (!vis[k]) {
				sg[j] = k;
				break;
			}
			add(f[i][sg[j]],h[j]);
		}
	}
mx=500;
	int ans = 0;
	REP(i,0,mx) REP(j,0,mx) {
		add(ans, (ll)f[0][i]*f[1][j]%P*f[2][i^j]);
	}
	printf("%d\n", ans);
}