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.util.*;
import java.io.*;
//import java.math.*;

public class Task{
        //  ..............code begins here..............

static long mod=(long)1e9+7,mod1=998244353l,inf=(long)1e17;

static void solve() throws IOException {
            int[] x=int_arr();
            int n=x[0],m=x[1];
            f=new long[25];
            f[0]=1;
            for(int i=1;i<25;i++){
                f[i]=Modmul(f[i-1],(long)i,mod1);
            }
            int[][] a=new int[m][n];
            for(int i=0;i<n;i++){
                int[] d=int_arr();
                for(int j=0;j<m;j++){
                    a[j][i]=d[j];
                }
            }
            long res=m;
            for(int i=0;i<m;i++){
                int[] fre=new int[n+2];
                for(int j=0;j<n;j++)fre[a[i][j]]++;
                long tmp=1l;
                int occ=0;
                for(int j=1;j<=n+1;j++){
                    tmp=Modmul(tmp,nck(j-1-occ,fre[j],mod1),mod1);
                    tmp=Modmul(tmp,f[fre[j]],mod1);
                    occ+=fre[j];
                }
                tmp=Modmul(tmp,ModInv(f[n],mod1),mod1);
                res-=tmp;
            }
            res%=mod1;
            res=(res+mod1)%mod1;
            out.write(res+"\n");
}
//
public static void main(String[] args) throws  IOException{
                assign();
                int t=1;//int_v(read()),cn=1;
                while(t--!=0){
                    //out.write("Case #"+cn+": ");
                    solve();
                    //cn++;
                }
                out.flush();
}

// taking inputs
static BufferedReader s1;
static BufferedWriter out;
static String read() throws IOException{String line="";while(line.length()==0){line=s1.readLine();continue;}return line;}
static int int_v (String s1){return Integer.parseInt(s1);}
static long long_v(String s1){return Long.parseLong(s1);}
static void sort(int[] a){List<Integer> l=new ArrayList<>();for(int x:a){l.add(x);}Collections.sort(l);for(int i=0;i<a.length;i++){a[i]=l.get(i);}}
static int[] int_arr() throws IOException{String[] a=read().split(" ");int[] b=new int[a.length];for(int i=0;i<a.length;i++){b[i]=int_v(a[i]);}return b;}
static long[] long_arr() throws IOException{String[] a=read().split(" ");long[] b=new long[a.length];for(int i=0;i<a.length;i++){b[i]=long_v(a[i]);}return b;}
static void assign(){s1=new BufferedReader(new InputStreamReader(System.in));out=new BufferedWriter(new OutputStreamWriter(System.out));}
//static String setpreciosion(double d,int k){BigDecimal d1 = new BigDecimal(Double.toString(d));return d1.setScale(k,RoundingMode.HALF_UP).toString();}//UP DOWN HALF_UP
static int add(int a,int b){int z=a+b;if(z>=mod)z-=mod;return z;}
static long gcd(long a,long b){if(b==0){return a;}return gcd(b,a%b);}
static long Modpow(long a,long p,long m){long res=1;while(p>0){if((p&1)!=0){res=(res*a)%m;}p >>=1;a=(a*a)%m;}return res%m;}
static long Modmul(long a,long b,long m){return ((a%m)*(b%m))%m;}
static long ModInv(long a,long m){return Modpow(a,m-2,m);}
static long nck(int n,int r,long m){if(r>n){return 0l;}return Modmul(f[n],ModInv(Modmul(f[n-r],f[r],m),m),m);}
static long[] f;
}