#include "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = 1'000'000'007;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }

// 拡張ユークリッドの互除法
// 一次不定方程式 ax + by = gcd(a, b) を満たす x, y を求める
// ax + by = k * gcd(a, b) は、まず ax + by = gcd(a, b) を解き、解を k 倍する
// 戻り値: gcd(a, b)
long long extgcd(long long a, long long b, long long &x, long long &y) {
	long long g = a; x = 1; y = 0;
	if (b != 0) {
		g = extgcd(b, a % b, y, x);
		y -= (a / b) * x;
	}
	return g;
}

// 中国剰余定理
pair<long long, long long> crt(const vector<long long> &a, const vector<long long> &n) {
	assert(a.size() == n.size());
	long long A = 0, N = 1;
	for (int i = 0; i < a.size(); i++) {
		long long u, v;
		long long g = extgcd(N, n[i], u, v);
		if (a[i] % g != A % g)
			return make_pair(-1, -1);
		long long m = n[i] / g;
		long long t = (a[i] - A) / g * u % m;
		A += N * t;
		N *= m;
	}
	A %= N;
	if (A < 0)A += N;
	return make_pair(A % N, N);
}

int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
int lcm(int x, int y) { return x / gcd(x, y) * y; }

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, K; cin >> N >> K;
	vector<int> X(K), Y(K); rep(i, 0, K) {
		cin >> X[i] >> Y[i];
		X[i]--, Y[i]--;
	}
	vector<int> v(N); iota(all(v), 0);
	rep(i, 0, K)
		swap(v[X[i]], v[Y[i]]);
	int Q; cin >> Q;
	vector<vector<int>> A(Q, vector<int>(N));
	rep(i, 0, Q) rep(j, 0, N) { cin >> A[i][j]; A[i][j]--; }
	vector<int> g(N);
	vector<vector<int>> r(N, vector<int>(N, -1));
	int l = 1;
	rep(i, 0, N) {
		int c = 1;
		r[i][i] = 0;
		for (int s = v[i]; s != i; s = v[s]) {
			r[i][s] = c;
			c++;
		}
		g[i] = c;
		l = lcm(l, c);
	}
	dump(r);
	rep(q, 0, Q) {
		vector<long long> a, n;
		rep(i, 0, N) {
			a.push_back(r[i][A[q][i]]);
			n.push_back(g[i]);
		}
		if(count(all(a),-1))
			cout << -1 << endl;
		else {
			auto res = crt(a, n);
			if(res.first == -1)
				cout << -1 << endl;
			else if(res.first == 0)
				cout << l << endl;
			else
				cout << res.first << endl;
		}
	}
	return 0;
}