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}
//https://github.com/EgorKulikov/yaal/tree/master/lib/main/net/egork
        
import java.util.*;
import java.io.*;
public class A{
    static PrintWriter out;
    static InputReader in;
    public static void main(String args[]){
        out = new PrintWriter(System.out);
        in = new InputReader();
        new A();
        out.flush(); out.close();
    }   
    A(){
        solve();
    }
    final int mod = 998244353;
    final int maxn = 22;
    long fac[] = new long[maxn], ifac[] = new long[maxn];
    long mul(long a, long b){
        a *= b;
        if(a >= mod)a %= mod;
        return a;
    }
    long add(long a, long b){
        a += b;
        if(a >= mod)a -= mod;
        return a;
    }
    long sub(long a, long b){
        a -= b;
        if(a < 0)a += mod;
        return a;
    }
    long rpe(long a, long b){
        long ans = 1;
        while(b != 0){
            if(b % 2 == 1)ans = mul(ans, a);
            a = mul(a, a); b >>= 1;
        }return ans;
    }
    long ncr(int n, int r){
        if(r > n)return 0;
        return mul(fac[n], mul(ifac[r], ifac[n - r]));
    }
    long npr(int n, int r){
        if(r > n || n < 0)return 0;
        return mul(fac[n], ifac[n - r]);
    }
    void solve(){
        // int t = in.nextInt();
        fac[0] = ifac[0] = 1;
        for(int i = 1; i < maxn; i++){
            fac[i] = mul(fac[i - 1], i);
            ifac[i] = rpe(fac[i], mod - 2);
        }
        int t = 1;
        while(t-- > 0){
            int n = in.nextInt(), m = in.nextInt();
            int a[][] = new int[n][m];
            for(int i = 0; i < n; i++){
                for(int j = 0; j < m; j++)a[i][j] = in.nextInt();
            }
            int dp[] = new int[maxn];
            long sub = 0;
            for(int i = 0; i < m; i++){
                Arrays.fill(dp, 0);
                for(int j = 0; j < n; j++)dp[a[j][i]]++;
                int space = 0;
                long cont = 1;
                for(int j = 1; j <= n + 1; j++){
                    if(dp[j] != 0){
                        cont = mul(cont, npr(space, dp[j]));
                        space -= dp[j];
                    }
                    space++;
                }
                // System.out.println(cont);
                sub = add(sub, cont);
            }
            long ans = sub(m, mul(sub, ifac[n]));
            out.print(ans);
        }
    }
    public static class InputReader{
        BufferedReader br;
        StringTokenizer st;
        InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public int nextInt(){
            return Integer.parseInt(next());
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
        public String next(){
            while(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){}
            }
            return st.nextToken();
        }
    }
}
	