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;
#define endl '\n'
#define ll long long
#define pll pair<ll,ll>
#define rep(i,n) for(ll i=0;i<n;i++)
#define mod 998244353 
#define INF 10000000000000000
#define F first
#define S second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define pie 3.141592653589793238462643383279
#define PYES cout<<"YES"<<endl
#define PNO cout<<"NO"<<endl
#define SB(a) sort(a.begin(),a.end(),greater<ll>());
#define SS(a) sort(a.begin(),a.end());
#define vll vector<ll>
#define vpll vector<pll>
vector<bool> prime;
vector<ll> fact,inv,primes,factors;
void fast()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
}
void factorize(ll a)
{
	factors.clear();
	for(ll i=1;i*i<=a;i++)
	{
		if (i*i==a) factors.pb(i);
		else if (a%i==0)
		{
			factors.pb(i);
			factors.pb(a/i);
		}
	}
	sort(factors.begin(),factors.end());
}
ll power(ll a,ll b)
{
	if(a==1||b==0)	return 1;
	ll c=power(a,b/2);
	ll res=1;
	res=c*c;
	if(res>=mod)	res%=mod;
	if(b%2)	res*=a;
	if(res>=mod)	res%=mod;
	return res;
}
ll modInv(int a)
{
	return power(a,mod-2);
}
void factorial(ll n)
{
	fact.resize(n+1);
	fact[0]=1;
	for(ll i=1;i<=n;i++)
	{
		fact[i]=fact[i-1]*i;
		if(fact[i]>=mod)	fact[i]%=mod;
	}
}
void InvFactorial(ll n)
{
	inv.resize(n+1);
	inv[0]=1;
	for(ll i=1;i<=n;i++)	inv[i]=modInv(fact[i]);
}
ll ncr(ll n,ll r)
{
	if(n<r||n<0||r<0)	return 0;
	ll b=inv[n-r];
	ll c=inv[r];
	ll a=fact[n]*b;
	if(a>=mod)	a%=mod;
	a*=c;
	if(a>=mod)	a%=mod;
	return a;
}
void remove_duplicates(vector<pair<ll,ll>> &v)
{
	sort(v.begin(),v.end());
	ll _size=unique(v.begin(),v.end())-v.begin();
	v.resize(_size);
}
unsigned ll gcd(unsigned ll u, unsigned ll v)
{
    if(u==0||v==0) return max(u,v);
    unsigned ll shift=__builtin_ctz(u|v);
	u>>=__builtin_ctz(u);
    do{
    	v>>=__builtin_ctz(v);
    	if(u>v)	swap(u,v);
    	v-=u;
    }while(v!=0);
    return u<<shift;
}
void sieve(int n)
{
    prime.assign(n+1,1);
	prime[1]=false;
    for(ll p=2;p*p<=n;p++) if(prime[p])	for(ll i=p*p;i<=n;i+=p)	prime[i]=false;
    for(ll i=2;i<n+1;i++) if(prime[i]) primes.push_back(i);
}
ll size(ll l)
{
	ll p=l;
	ll count=0;
	while(p) p/=10,count++;
	return count;
}
//-------------------------------------------------------------------------------------------------------------------------------------------------
// Always remember: if (condition) if (condition) statement; This doesn't work !!
// Always increase boundaries for global or frequency questions
// pow((ll)10,i) doesn't work sometimes. It is better to use multiple of 10 in arrays. Always remember this.
// delete values when defining global if there are multiple testcases
// don't try to be smart and make stupid mistakes
void solve()
{
	ll n,m;
	cin>>n>>m;
	vector<vll> d(n,vll(m));
	rep(i,n) rep(j,m) cin>>d[i][j];
	ll ans=0;
	factorial(n);
	InvFactorial(n);
	for(ll j=0;j<m;j++)
	{
		ll count=1;
		vll nb;
		for(ll i=0;i<n;i++) nb.pb(d[i][j]);
		SS(nb);
		for(ll i=0;i<n;i++) count*=nb[i]-i-1;
		ans+=fact[n]-count;
		if (ans<0)
		{
			ll negans=-ans;
			negans=(negans+mod-1)/mod;
			ans+=negans*mod;
		}
		ans%=mod;
	}
	cout<<(ans*inv[n])%mod;
}
signed main()
{
	fast();
	ll T=1;
	//cin>>T;
	for(ll i=1;i<=T;i++) solve();
}
