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//
// Created by piotr on 12.03.2024.
//
#include <array>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <numeric>
#include <set>
#include <vector>
#include <tuple>

struct Ułamek {
public:
	long long p;
	long long q;

	Ułamek(long long p=0, long long q=1) : p(p), q(q) {
		norm();
	}

	Ułamek& operator+=(const Ułamek& inny) {
		long long d = std::gcd(q, inny.q);
		p = (p * inny.q + inny.p * q) / d;
		q *= inny.q / d;
		norm();
		return *this;
	}

	Ułamek& operator-=(const Ułamek& inny) {
		long long d = std::gcd(q, inny.q);
		p = (p * inny.q - inny.p * q) / d;
		q *= inny.q / d;
		norm();
		return *this;
	}

private:
	void norm() {
		long long d = std::gcd(std::abs(p), q);
		p /= d;
		q /= d;
	}
};

std::tuple<long long, long long, long long> extended_gcd(long long a, long long b) {
	if (b == 0) {
		return {1, 0, a};
	}
	auto [x1, y1, gcd] = extended_gcd(b, a % b);
	return {y1, x1 - (a / b) * y1, gcd};
}

long long mod_inverse(long long a, long long m) {
	long long x, y, gcd;
	std::tie(x, y, gcd) = extended_gcd(a, m);
	if (gcd != 1) {
		return -1;
	}
	return (x % m + m) % m;
}

using Perm = std::vector<int>;

bool is_identity(const Perm& p) {
	const int N = p.size();
	for (int i=0; i<N; ++i) {
		if (p[i] != i) {
			return false;
		}
	}
	return true;
}

const int BLOCKS = 50;
const long long MODULO = 1000000007;

long long count_inversions(std::vector<int>& counts, std::vector<int>& countsB, const Perm& p) {
	const int N = p.size();
	const int maxB = N/BLOCKS;
	std::fill(counts.begin(), counts.end(), 0);
	std::fill(countsB.begin(), countsB.end(), 0);

	long long inversions = 0;
	for (int i=0; i<N; ++i) {
		// chcemy znaleźć liczbę elementów w counts pomiędzy p[i] a N-1
		int piB = p[i] / BLOCKS;
		int endC = std::min(BLOCKS*(piB+1), N);
		for (int j=p[i]; j<endC; ++j) {
			inversions += counts[j];
		}
		for (int j=piB+1; j<=maxB; ++j) {
			inversions += countsB[j];
		}

		counts[p[i]] += 1;
		countsB[piB] += 1;
	}
	return inversions;
}

void combine(Perm& p, const Perm& q) {
	for (auto& index : p) {
		index = q[index];
	}
}

void remove_duplicates(const Perm& p, std::set<Perm>& other_permutations) {
	Perm pn(p);
	other_permutations.erase(pn);
	do {
		combine(pn, p);
		other_permutations.erase(pn);
	}
	while (!is_identity(pn));
}

Ułamek count_inversions_simple(int N, const Perm& p)
{
	Perm pn(p);
	long long P=0, Q=0;
	std::vector<int> counts(N), countsB(N/BLOCKS+1);

	P = (P+count_inversions(counts, countsB, p))%MODULO;
	Q++;

	do {
		combine(pn, p);

		P = (P+count_inversions(counts, countsB, pn))%MODULO;
		Q++;
	}
	while (!is_identity(pn));

	return { P, Q % MODULO };
}

Ułamek count_inversions_smart(int N, const Perm& p)
{
	Ułamek u;
	std::vector<int> cycle_number(N, -1);

	int C = 0;
	for (int i0=0; i0<N; ++i0) {
		if (cycle_number[i0] == -1) {
			int i = i0;
			do {
				cycle_number[i] = C;
				i = p[i];
			} while (i != i0);
			++C;
		}
	}

	std::vector<Ułamek> poj(C);
	for (int c=0; c<C; ++c) {
		Perm pc(N);
		for (int i=0; i<N; ++i) {
			pc[i] = (cycle_number[i] == c) ? p[i] : i;
		}
		poj[c] = count_inversions_simple(N, pc);
		u += poj[c];
	}

	for (int c1=0; c1<C; ++c1) {
		for (int c2=c1+1; c2<C; ++c2) {
			Perm pc(N);
			for (int i=0; i<N; ++i) {
				pc[i] = (cycle_number[i] == c1 || cycle_number[i] == c2) ? p[i] : i;
			}
			u += count_inversions_simple(N, pc);
			u -= poj[c1];
			u -= poj[c2];
		}
	}

	return u;
}

int foo(int N, int K)
{
	std::vector<Perm> permutations;
	permutations.reserve(K);

	for (int k = 0; k<K; ++k) {
		Perm p(N);
		for (int n = 0; n<N; ++n) {
			int x;
			assert(scanf("%d", &x)==1);
			p[n] = x-1;
		}
		if (!is_identity(p)) {
			permutations.push_back(p);
		}
	}
	if (permutations.empty()) {
		return 0;
	}
	K = permutations.size(); // wszystkie Id odrzucamy

	Perm first_permutation = permutations[0];
	std::set<Perm> other_permutations(permutations.begin()+1, permutations.end());
	remove_duplicates(first_permutation, other_permutations);
	if (!other_permutations.empty()) {
		return N; // tak, wiem, że to nie do końca tak
	}

	auto pair = count_inversions_smart(N, first_permutation);
	long long P = pair.p, Q = pair.q;
	long long PQ = std::gcd(P, Q);
	P /= PQ;
	Q /= PQ;
	Q = mod_inverse(Q, MODULO);

	return (P * Q) % MODULO;
}

int main()
{
	int N, K;
	assert(scanf("%d%d", &N, &K) == 2);
	printf("%d\n", foo(N, K));
}