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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const int mod = 998244353;

long long int mpp(long long int n, long long int k, long long int p)
{
	long long int r = 1;
	while(k)
	{
		if(k & 1)
		{
			r = (r * n) % p;
		}
		k >>= 1;
		n = (n * n) % p;
	}
	return r;
}

int main()
{
	int n, m;
	cin >> n >> m;
	vector<vector<int>> A(n, vector<int>(m));
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
		{
			cin >> A[i][j];
		}
	}
	
	long long int d = 1;
	for(int i = 2; i <= n; i++)
	{
		d *= i;
		d %= mod;
	}
	d = mpp(d, mod-2, mod);
	
	long long int total = 0;
	
	for(int i = 0; i < m; i++)
	{
		long long int q = 1; // probability that current point is not covered
		vector<int> F(n);
		for(int j = 0; j < n; j++)
		{
			F[j] = A[j][i];
		}
		sort(F.begin(), F.end());
		int filled = 0;
		for(int j = 0; j < n; j++)
		{
			int h = F[j]-1;
			if(h > filled)
			{
				q *= (h-filled);
				//q %= mod;
				filled++;
			}
			else
			{
				q = 0;
				break;
			}
		}
		
		q %= mod;
		q = (q * d) % mod;
		long long int p = (1 + (mod - q));
		
		total += p;
	}
	total %= mod;
	total = (mod + total) % mod;
	cout << total << endl;

	return 0;
	
}
	
	
