Monocarp is playing a game "Assimilation IV". In this game he manages a great empire: builds cities and conquers new lands.

Monocarp's empire has n cities. In order to conquer new lands he plans to build one Monument in each city. The game is turn-based and, since Monocarp is still amateur, he builds exactly one Monument per turn.

Monocarp has m points on the map he'd like to control using the constructed Monuments. For each point he knows the distance between it and each city. Monuments work in the following way: when built in some city, a Monument controls all points at distance at most 1 to this city. Next turn, the Monument controls all points at distance at most 2, the turn after — at distance at most 3, and so on. Monocarp will build n Monuments in n turns and his empire will conquer all points that are controlled by at least one Monument.

Monocarp can't figure out any strategy, so during each turn he will choose a city for a Monument randomly among all remaining cities (cities without Monuments). Monocarp wants to know how many points (among m of them) he will conquer at the end of turn number n. Help him to calculate the expected number of conquered points!

Input

The first line contains two integers n and m (1 ≤ n ≤ 20; 1 ≤ m ≤ 5 ⋅ 10^4) — the number of cities and the number of points.

Next n lines contains m integers each: the j-th integer of the i-th line d_{i, j} (1 ≤ d_{i, j} ≤ n + 1) is the distance between the i-th city and the j-th point.

Output

It can be shown that the expected number of points Monocarp conquers at the end of the n-th turn can be represented as an irreducible fraction x/y. Print this fraction modulo 998 244 353, i. e. value x ⋅ y^{-1} mod 998244353 where y^{-1} is such number that y ⋅ y^{-1} mod 998244353 = 1.

Example

Input


3 5
1 4 4 3 4
1 4 1 4 2
1 4 4 4 3


Output


166374062

Note

Let's look at all possible orders of cities Monuments will be build in: 

  * [1, 2, 3]: 
    * the first city controls all points at distance at most 3, in other words, points 1 and 4; 
    * the second city controls all points at distance at most 2, or points 1, 3 and 5; 
    * the third city controls all points at distance at most 1, or point 1. 
In total, 4 points are controlled. 
  * [1, 3, 2]: the first city controls points 1 and 4; the second city — points 1 and 3; the third city — point 1. In total, 3 points. 
  * [2, 1, 3]: the first city controls point 1; the second city — points 1, 3 and 5; the third city — point 1. In total, 3 points. 
  * [2, 3, 1]: the first city controls point 1; the second city — points 1, 3 and 5; the third city — point 1. In total, 3 points. 
  * [3, 1, 2]: the first city controls point 1; the second city — points 1 and 3; the third city — points 1 and 5. In total, 3 points. 
  * [3, 2, 1]: the first city controls point 1; the second city — points 1, 3 and 5; the third city — points 1 and 5. In total, 3 points. 

The expected number of controlled points is (4 + 3 + 3 + 3 + 3 + 3)/(6) = 19/6 or 19 ⋅ 6^{-1} ≡ 19 ⋅ 166374059 ≡ 166374062 \pmod{998244353}
#include <algorithm>    // std::sort
#include <iostream>
#include <string>		// to_string
#include <vector>
#include <unordered_map>
#include <map>
#include <unordered_set>
#include <set>
#include <queue>
#define ll long long
using namespace std;

int mod = 998244353;
vector<int> inverseH(int a, int b) {
	if (a == 1) {
		// cout << "{1, " << to_string(b) << "} -> {1, 0} \n";
		return vector<int>{ 1, 0 };
	}
	if (a > b) {
		vector<int> temp = inverseH(b, a);
		vector<int> res;
		res.push_back(temp.at(1));
		res.push_back(temp.at(0));
		return res;
	}
	else {
		vector<int> temp = inverseH(b % a, a);
		// temp(1) * a + temp(0) * (b % a) = 1
		// (temp(1) - temp(0)*(b-b%a)/a) * a + temp(0) * b = 1
		vector<int> res;
		int num = (b - b % a) / a;
		num = (int)((long long)(temp.at(0)) * num);
		num = (temp.at(1) - num);
		res.push_back(num);

		res.push_back(temp.at(0));
		// cout << "{" << to_string(a) << ", " << to_string(b) << "} -> {" << to_string(res.at(0)) << ", " << to_string(res.at(1)) << "} \n";
		return res;
	}
}

// Req: inverseH, src: 1493D
int inverse(int a, int b) {
	int val = inverseH(a, b).at(0);
	val = val % b;
	if (val < 0) {
		val += b;
	}
	return val;
}
void solve() {
	// string:
	//char** vals = new char* [n];
	//for (int i = 0; i < n; i++) {
	//	string str = "";
	//	cin >> str;
	//	vals[i] = new char[n];
	//	for (int j = 0; j < n; j++) {
	//		vals[i][j] = str[j];
	//	}
	//}
	int n, m;
	cin >> n >> m;
	int** vals = new int* [m];
	for (int i = 0; i < m; i++) {
		vals[i] = new int[n];
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			cin >> vals[j][i];
		}
	}
	ll fact = 1;
	for (int i = 1; i <= n; i++) {
		fact *= i;
		fact = fact % mod;
	}
	int invFact = inverse((int)fact, mod);
	ll ans = 0;
	for (int i = 0; i < m; i++) {
		ll num = 1;
		for (int j = n; j > 0; j--) {
			int cnt = 0;
			for (int k = 0; k < n; k++) {
				if (vals[i][k] > j) {
					cnt++;
				}
			}
			num *= (cnt-(n-j));
			num = num % mod;
		}
		num = (mod + fact - num) % mod;
		// cout << "NUM: " << num << "\n";
		num *= invFact;
		num = num % mod;
		ans += num;
		ans = ans % mod;
	}
	cout << ans;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	solve();
}