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:
// I think of what the world could be, A vision of the one I see, A million dreams is all it's gonna take

#include<bits/stdc++.h>

#define F first
#define S second
#define PB push_back
#define sz(s) int((s).size())
#define bit(n,k) (((n)>>(k))&1)

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

const int maxn = 1e5 + 10, mod = 998244353;
const int MX = 1 << 17;

int cnt1[MX], cnt2[MX], cnt3[MX], arr[MX];
int n, dp[maxn];
vector<int> v[maxn];

int Pow(int a, int b){
    int ans = 1;
    for(; b; b >>= 1, a = 1ll * a * a % mod)
	if(b&1)
	    ans = 1ll * ans * a % mod;
    return ans;
}
void solve(int *arr){
    int m;
    cin >> m;
    for(int i = 1; i <= n; i++){
	v[i].clear();
    }
    while(m--){
	int a, b;
	cin >> a >> b;
	if(a > b)
	    swap(a, b);
	v[a].PB(b);
    }
    for(int i = n; i >= 1; i--){
	vector<int> vec;
	for(int j : v[i])
	    vec.PB(dp[j]);
	sort(vec.begin(), vec.end());
	vec.resize( unique(vec.begin(), vec.end()) - vec.begin() );
	dp[i] = sz(vec);
	for(int j = 0; j < sz(vec); j++){
	    if(j != vec[j]){
		dp[i] = j;
		break;
	    }
	}
	assert(dp[i] < MX);
	arr[dp[i]] = (arr[dp[i]] + Pow(10, 18 * i)) % mod;
    }
}
void txor(int *a,int tw,bool inv){
    for(int ln=1;(2*ln)<=tw;ln*=2){
	for(int l=0;l<tw;l+=ln+ln){
	    for(int i=0;i<ln;i++){
		int x=a[l+i],y=a[l+ln+i];
		a[l+i]=(x+y) % mod;
		a[l+ln+i]=(x-y) % mod;
	    }
	}
    }
    if(inv){
	int r = Pow(tw, mod-2);
	for(int i=0;i<tw;i++)
	    a[i] = 1ll * a[i] * r % mod;
    }
}

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

    cin >> n;
    solve(cnt1), solve(cnt2), solve(cnt3);

    txor(cnt1, 1<<17, 0);
    txor(cnt2, 1<<17, 0);
    for(int i = 0; i < (1<<17); i++)
	cnt1[i] = 1ll * cnt1[i] * cnt2[i] % mod;
    txor(cnt1, 1<<17, 1);
    int ans = 0;
    for(int i = 0; i < (1<<17); i++)
	ans = (1ll * ans + 1ll * cnt1[i] * cnt3[i]) % mod;
    if(ans < 0)
	ans+= mod;
    return cout << ans << endl, 0;
}