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}
/*
bts songs to dance to:
I need U
Run
ON
Filter
I'm fine
 */
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1525E
{
    static final long MOD = 998244353L;
    public static void main(String hi[]) throws Exception
    {
        fac = new long[22];
        invfac = new long[22];
        fac[0] = invfac[0] = 1L;
        for(int i=1; i < 22; i++)
        {
            fac[i] = (fac[i-1]*i)%MOD;
            invfac[i] = power(fac[i], MOD-2, MOD);
        }
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int N = Integer.parseInt(st.nextToken());
        int M = Integer.parseInt(st.nextToken());
        int[][] grid = new int[N][M];
        for(int r=0; r < N; r++)
            grid[r] = readArr(M, infile, st);
        long res = 0L;
        for(int a=0; a < M; a++)
        {
            cnt = new int[N];
            for(int i=0; i < N; i++)
                if(grid[i][a] > 1)
                    cnt[N+1-grid[i][a]]++;
            prefix = new int[N];
            prefix[0] = cnt[0];
            for(int i=1; i < N; i++)
                prefix[i] = prefix[i-1]+cnt[i];
            dp = new long[N];
            Arrays.fill(dp, -1);
            long temp = dfs(0, N);
            temp = (temp*invfac[N])%MOD;
            temp = (1-temp+MOD)%MOD;
            res += temp;
            if(res >= MOD)
                res -= MOD;
        }
        System.out.println(res);
    }
    static int[] cnt;
    static int[] prefix;
    static long[] dp;
    static long[] fac, invfac;
    public static long dfs(int i, int N)
    {
        if(i == N)
            return 1L;
        if(dp[i] != -1)
            return dp[i];
        if(prefix[i]-i <= 0)
            return dp[i] = 0L;
        long ways = prefix[i]-i;
        return dp[i] = (ways*dfs(i+1, N))%MOD;
    }
    public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
    {
        int[] arr = new int[N];
        st = new StringTokenizer(infile.readLine());
        for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
        return arr;
    }
    public static long power(long x, long y, long p)
    {
        //0^0 = 1
        long res = 1L;
        x = x%p;
        while(y > 0)
        {
            if((y&1)==1)
                res = (res*x)%p;
            y >>= 1;
            x = (x*x)%p;
        }
        return res;
    }
}