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}
#include <bits/stdc++.h>
  
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define fori(i,n) for (int i=0;i<n;++i)
#define ford(i,n) for (int i=n-1;i>=0;--i)
#define forskip(i,n) for (int i=1;i<n;++i)
#define each(v) v.begin(), v.end()
#define pb push_back
#define mp make_pair
ll mod=998244353;

ll pow(ll x, ll y){
	ll res=1;
	x=x%mod;
	while (y>0){
		if (y&1){
			res=(res*x)%mod;
		}
		y=y>>1;
		x=(x*x)%mod;
	}
	return res;
}

ll mi(ll n){
	return pow(n,mod-2);
}

ll modfac(ll n){
	if (n==0) return 1;
	return (n*modfac(n-1))%mod;
}

ll nCr(ll n, ll r){
	if (n<r) return 0;
	if (r==0) return 1;
	return (modfac(n)*mi(modfac(r)) % mod
            * mi(modfac(n - r)) % mod)%mod;
}

void fast(){
        ios_base::sync_with_stdio(0);
        cin.tie(NULL);
        cout.tie(NULL);
}

int main(){
        fast();
        int n,m; cin >> n >> m;
	ll d[n][m]; fori(i,n) fori(j,m) cin >> d[i][j];
	vector<int> dp[m];
	fori(i,n) fori(j,m) dp[j].pb(d[i][j]);
	fori(i,m) sort(each(dp[i]));
	ll ans=0;
	ll pre=mi(modfac(n))%mod;
	ll p=modfac(n)%mod;
//	cout << pre << '\n';
	fori(i,m){
		ll t=0;
		ll m=1;
		fori(j,n){
			t=dp[i].end()-upper_bound(dp[i].begin(),dp[i].end(),n-j)-j;
	//		cout << t << " " << dp[i][j] << '\n';
			m=(m*max(t,0LL))%mod;
		}
		m=(p-m+mod)%mod;
	//	cout << m << '\n';
		ans=(ans+m)%mod;
	}
	cout << (ans*pre)%mod << '\n';
}
