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;
#define rep(i,a,b) for(int i=a;i<b;++i)
#define debug(x) cerr<<#x<<": "<<x<<endl
#define sz(x) (int)(x).size()
#define trav(a,x) for(auto &a:x)
#define all(x) x.begin(), x.end()
using ll = long long;
const ll MOD=998244353;
vector<int> labels(vector<vector<int>> g) {
    int n = sz(g);
    vector<int> label(n);
    for(int i=n-1;i>=0;--i) {
        vector<int> others;
        for(int j:g[i]) if(j>i) others.push_back(label[j]);
        sort(all(others));
        int k=0;
        for(int x:others) {
            if(x==k) ++k;
            else if(x>k) break;
        }
        label[i]=k;
    }
    return label;
}
vector<vector<int>> getgraph(int n) {
    int m;
    cin>>m;
    vector<vector<int>> g(n);
    rep(i,0,m) {
        int a,b;
        cin>>a>>b;
        --a;--b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    return g;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin>>n;
    ll B = 1;
    rep(i,0,18) B=(B*10)%MOD;
    vector<ll> power(n,B);
    rep(i,1,n) power[i]=(B*power[i-1])%MOD;
    auto g1 = getgraph(n);
    auto g2 = getgraph(n);
    auto g3 = getgraph(n);
    auto l1 = labels(g1), l2 = labels(g2), l3=labels(g3);
    vector<ll> w1(2*n), w2(2*n), w3(2*n);
    rep(i,0,n) {
        w1[l1[i]]=(w1[l1[i]]+power[i])%MOD;
        w2[l2[i]]=(w2[l2[i]]+power[i])%MOD;
        w3[l3[i]]=(w3[l3[i]]+power[i])%MOD;
    }
    // rep(i,0,n) {
    //     cerr<<w1[i]<<" "<<w2[i]<<" "<<w3[i]<<"\n";
    // }
    int mxl1=0,mxl2=0;
    rep(i,0,n) {
        if(w1[i]!=0) mxl1=i;
        if(w2[i]!=0) mxl2=i;
    }
    ll ans=0;
    rep(i,0,mxl1+1) rep(j,0,mxl2+1) {
        ll x=(w1[i]*w2[j])%MOD;
        ans=(ans+x*w3[i^j])%MOD;
    }
    cout<<ans<<endl;
}
