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 ll long long
#define f(i,a,b) for(ll i=a;i<=b;i++)
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,avx,mmx,tune=native")
#define fi first
#define se second

ll n,m,d[22][50005],mod=998244353;

ll pang(ll x,ll y){
	if(x==0)return 0;
	if(y==0)return 1;
	if(y==1)return x;
	ll z=pang(x,y/2);
	return ((z*z)%mod*pang(x,y%2))%mod;
}

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n>>m;
	f(i,1,n){
		f(j,1,m){
			cin>>d[i][j];
		}
	}
	ll bum=1;
	f(i,1,n){
		bum=(bum*i)%mod;
	}
	ll sum=0;
	f(i,1,m){
		ll cnt=1;
		vector<ll>v;
		v.clear();
		f(j,1,n){
			ll temp=d[j][i]-1;
			v.push_back(temp);
		}
		sort(v.begin(),v.end());
		f(j,0,n-1){
			ll temp=max((ll)0,v[j]-j);
			cnt=(cnt*temp)%mod;
		}
		sum=(sum+bum-cnt)%mod;
	}
	sum=(sum+2*mod)%mod;
	ll kali=pang(bum,mod-2);
	cout<<(sum*kali)%mod<<'\n';
}
