#include<bits/stdc++.h>
typedef long long ll;
#define vll vector<ll>
#define pb push_back
#define pll pair<ll,ll>
#define vpll vector<pair<ll,ll> >
#define st(x) sort(x.begin(),x.end())
#define sz(x) (ll)x.size()
#define input(n,x) for(ll i=0;i<n;i++){ll k;cin>>k;x.pb(k);}
#define fr first
#define sc second
#define rep(i,j,k) for(ll i=j;i<k;i++)
#define repn(i,j,k) for(ll i=j;i>=k;i--)
using namespace std;
vll x;
const ll mod =  998244353;
ll inv(ll a, ll b) {
	if (b == 0) {
		return 1;
	}
	if (b == 1) {
		return a;
	}
	ll k = inv(a, b / 2);
	k = (k * k) % mod;
	if (b & 1) {
		return (k * a) % mod;
	}
	return k;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	// IF USE OF LEAST / MIN  ANYWHERE IN THE STATEMENT FIRST THOUGHT SHOULD BE "BINARY SEARCH"
	// THINK DP WHENEVER N*M<=1000000 AT MAX
	// WRITE DOWN EVERYTHING
	// FOR FLOWS USE DINICS ALGORITHM , TEMPLATE IS PRESENT IN ROCK PAPER SCISSOR PROBLEM
	// CHECK FOR GCD , LCM APPROACH . HARD TO SPOT
	// for problems related to subsequence the best dp is - the number of  subsequences that end at index i and have length j is equal to addition of the number of subsequences that end at index (0 ... i-1) and hve length j-1

	ll n, m;
	cin >> n >> m;
	ll b[n + 1][m + 1];
	rep(j, 0, n) {
		rep(k, 0, m) {
			cin >> b[j][k];
		}
	}
	ll f = 1;
	rep(j, 2, n + 1) {
		f = (f * j) % mod;
	}
	ll ans = 0;
	rep(j, 0, m) {
		vll pos;
		rep(k, 0, n) {
			pos.pb(min(n, b[k][j] - 1));
		}
		st(pos);
		ll t = 1;
		rep(j, 0, n) {
			t = (t * max(ll(0), pos[j] - j)) % mod;
		}
		ans = (ans + (f - t + mod) % mod + mod) % mod;
	}
	cout << (ans * inv(f, mod - 2)) % mod << "\n";
	return 0;
}