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}
// Author: Pawan Raj
// Problem: E. Assimilation IV
// Contest: Codeforces - Educational Codeforces Round 109 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1525/problem/E

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define int long long
#define F(i, n) for (int i = 0; i < n; i++)
#define all(x)(x).begin(), (x).end()
#define show(A) for (auto i: A) cout << i << " "; cout << '\n';
#define show_(A, n) F(i, n) { show(A[i]) }

using namespace std;
using ld = long double;
using vi = vector < int > ;
using mi = map < int, int > ;
using pi = pair < int, int > ;

const int N = 300005;
const int MOD = 998244353;
const int inf = 1e18 + 1;
/***************************code begins here*****************************/
int power(int x, unsigned int y) {  
    int res = 1;
    x = x % MOD;
    if (x == 0) return 0;
  
    while (y > 0)  
    { 
        if (y & 1)  
            res = (res*x) % MOD;
        y = y>>1;
        x = (x*x) % MOD;  
    }  
    return res;  
}

int fact(int n){
	if(n==0) return 1;
	return (n*fact(n-1))%MOD;
}
void solve() {
    int n, m, ans = 0; cin >> n >> m;
    
    int a[n][m];
    F(i, n) F(j, m) cin >> a[i][j];
    
    F(j, m){
    	int temp = 1;
    	vi c(n);
    	F(i, n){
    		c[i] = a[i][j]-1;
    	}
    	sort(all(c));
    	
    	// show(c);
    	
    	F(i, n){
    		if((min(n, c[i]) - i )<=0){
    			temp = 0;
    			break;
    		}
    		temp *= min(n, c[i]) - i;
    		temp %= MOD;
    	}
    	temp = (temp*power(fact(n),MOD-2))%MOD;
    	ans = (ans + MOD + 1 - temp)%MOD;
    }
    cout << ans << '\n';
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    // cin >> t;
    while (t--) solve();

    return 0;
}