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}
//int max = 2 147 483 647 (2^31-1)
//ll max = 9 223 372 036 854 775 807 (2^63-1)
#include<bits/stdc++.h>
using namespace std;

#define forn(i,n) for(int i=0;i<n;i++)
#define mp make_pair
#define f first
#define s second
#define pb push_back
#define MOD 998244353

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int,int> pi;

//Fast input and output
void fast_io(){
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);}

//Printing pairs and vectors
template<typename A, typename B> ostream& operator<< (ostream &cout, pair<A,B> const &p)
{return cout << "(" << p.f << ", " << p.s << ")";}
template<typename A> ostream& operator<< (ostream &cout, vector<A> const&v)
{cout << "["; forn(i,(int)v.size()){ if (i) cout << ", "; cout << v[i];} return cout << "]";}

ll mpow(ll base, ll exp) {
  ll res=1;
  while (exp) {
    if (exp%2==1) res=(res*base)%MOD;
    exp >>= 1; base=(base*base)%MOD;}
  return res;
}

ll minv(ll base) {
    return mpow(base, MOD - 2);
}

int main(){
  fast_io();
  int n,m; cin >> n >> m;
  ll k=1;
  forn(i,n) k=(k*(i+1))%MOD;
  k=minv(k);
  vector<vi>a(n,vi(m));
  forn(i,n) forn(j,m) cin >> a[i][j];
  ll ans=m;
  forn(j,m){
    bool mb=false;
    forn(i,n) if(a[i][j]==1) mb=true;
    if(mb) continue;
    vi p(n);
    forn(i,n) p[n+1-a[i][j]]++;
    ll w=1; int c=1;
    forn(i,n){
      c+=p[i]-1; w=(w*c)%MOD;}
    ans=((ans-w*k)%MOD+MOD)%MOD;
  }
  cout << ans;
}
//code by DanTheMan
