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  mp make_pair
#define REP(i,a,b) for(int i=a;i<b;i++)
typedef  long long LL ;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const ULL maxn = 5e4+10;
const ULL MOD = 998244353;
const int INF = 1e9+7;
const double PI = acos(-1);
const double eps = 1e-7;

ULL dis[25][maxn];
ULL n , m;
ULL fac[25];


void fact()
{
	ULL res = 1;
	fac[1]  = 1;
	fac[0] = 1;
	for(ULL i=1;i<=22;++i)
		fac[i] = ( i * fac[i-1] ) %MOD;
}



ULL qpow(ULL di, ULL mi)
{
	ULL res = 1;
	while(mi)
	{
		if(mi&1)
			res = (res *di)%MOD;
		di = (di*di)%MOD;
		mi>>=1;
	}
	return res;

}

ULL calc( int x)
{
	ULL a[25];
	for(int i=1;i<=n;++i)
		a[i] = dis[i][x];
	sort(a+1, a+1+n);
	ULL res = 1;
	ULL cnt  = 0;
	for(int i=1;i<=n;++i)
	{
		if(a[i]>n)
			cnt++;
		else res = (res * (a[i]-i))%MOD;
	}
	res = (res * fac[cnt] ) %MOD;
		//printf("%llu\n", ((fac[n]%MOD - res%MOD)+MOD)%MOD);
	return ((fac[n]%MOD - res%MOD)+MOD)%MOD;
}


void solve()
{
	fact();
	cin >> n >> m;
	for(int i=1;i<=n;++i)
	{
		for(int k=1;k<=m;++k)
			cin >> dis[i][k];
	}

	ULL ans = 0 ;
	for(int i=1;i<=m;++i)
		 ans =  (ans + calc(i))%MOD;

	ULL prime =  qpow(fac[n], MOD-2);
	printf("%llu", (ans * prime)%MOD );
}

int main()
{

	solve();
	return 0;
}
