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}
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class tr1 {
	static PrintWriter out;
	static StringBuilder sb;
	static long mod = (long) 998244353;
	static int[][] memo;
	static int n, m;
	static ArrayList<Integer>[] ad;
	static HashMap<Integer, Integer>[] going;
	static long inf = Long.MAX_VALUE;
	static char[][] g;
	static boolean[] vis;
	static int[] a;
	static int ans;
	static int[][] turn;
	static HashMap<Integer, Integer> hmC, hmG;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		n = sc.nextInt();
		m = sc.nextInt();
		int[][] a = new int[n][m];
		for (int i = 0; i < n; i++)
			a[i] = sc.nextArrInt(m);
		long ans = 0;
		long[] f = new long[n + 5];
		f[0] = 1;
		for (int i = 1; i < f.length; i++)
			f[i] = (f[i - 1] * i) % mod;
		for (int i = 0; i < m; i++) {
			Integer[] d = new Integer[n];
			for (int j = 0; j < n; j++)
				d[j] = a[j][i];
			Arrays.sort(d);
			if (d[0].intValue() == 1) {
				ans = (ans + f[n]) % mod;
				continue;
			}
			int all = 0;
			long mul = 1;
			for (int j = 0; j < n; j++) {
				if (d[j].intValue() > n)
					break;
				int x = d[j].intValue() - 1;
				mul = (mul * (x - all)) % mod;
				all++;
			}
			mul = (mul * f[n - all]) % mod;
			ans += (((f[n] - mul) % mod) + mod) % mod;
			ans %= mod;
		}

		System.out.println(ans * modPow(f[n], mod - 2) % mod);
		out.flush();
	}

	static long modPow(long a, long e)

	{
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1)
				res = (res * a) % mod;
			a = (a * a) % mod;
			e >>= 1;
		}
		return res;
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nextArrInt(int n) throws IOException {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public long[] nextArrLong(int n) throws IOException {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public int[] nextArrIntSorted(int n) throws IOException {
			int[] a = new int[n];
			Integer[] a1 = new Integer[n];
			for (int i = 0; i < n; i++)
				a1[i] = nextInt();
			Arrays.sort(a1);
			for (int i = 0; i < n; i++)
				a[i] = a1[i].intValue();
			return a;
		}

		public long[] nextArrLongSorted(int n) throws IOException {
			long[] a = new long[n];
			Long[] a1 = new Long[n];
			for (int i = 0; i < n; i++)
				a1[i] = nextLong();
			Arrays.sort(a1);
			for (int i = 0; i < n; i++)
				a[i] = a1[i].longValue();
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}