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}
//by szh
#include<bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pii pair<int,int>
#define pll pair<long long,long long>
#define pb push_back
#define debug(x) cerr<<#x<<"="<<x<<endl
#define pq priority_queue
#define inf 0x3f
#define rep(i,a,b) for (int i=a;i<(b);i++)
#define MP make_pair
#define SZ(x) (int(x.size()))
#define ll long long
#define mod 998244353
#define ALL(x) x.begin(),x.end()
void inc(int &a,int b) {a=(a+b)%mod;}
void dec(int &a,int b) {a=(a-b+mod)%mod;}
int lowbit(int x) {return x&(-x);}
ll p0w(ll base,ll p) {ll ret=1;while(p>0){if (p%2ll==1ll) ret=ret*base%mod;base=base*base%mod;p/=2ll;}return ret;}

const int maxn=5e4+10;

int n,m;
int grid[22][maxn];
int cnt[22];
int sum=0,tot=1;

int solve(int x) {
	memset(cnt,0,sizeof(cnt));
	rep(i,0,n) cnt[grid[i][x]]++;
	ll tmp=cnt[n+1];
	ll ret=1;
	rep(i,0,n) {
		if (tmp<=(ll)i) return 0;
		ret=ret*(tmp-i)%mod;
		tmp+=cnt[n-i];
	}
	return ret;
}

int main() {
//	freopen("input.txt","r",stdin);	
	std::ios::sync_with_stdio(false);cin.tie();
	cin>>n>>m;
	rep(i,2,n+1) tot=1ll*tot*i%mod;
	rep(i,0,n)
		rep(j,0,m) cin>>grid[i][j];
	rep(i,0,m) inc(sum,(tot-solve(i)+mod)%mod);
	cout<<1ll*sum*p0w(tot,mod-2)%mod;
	return 0;
}
