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 <cstring>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <string>
#include <map>
#include <iostream>
#include <vector>
#include <queue>
#include <unordered_map>
#include <random>
#include <stack>
#include <set>
#include <list>
#include <unordered_set>
//#include <atcoder/convolution>
//#include <atcoder/modint>

#define bug(x) cout<<"zdongdebug1: "<<x<<endl;
#define bug2(x, y) cout<<"zdongdebug2: "<<x<<" "<<y<<endl;
#define bug3(x, y, z) cout<<"zdongdebug3: "<<x<<" "<<y<<" "<<z<<endl;
#define bug4(x, y, z, w) cout<<"zdongdebug4: "<<x<<" "<<y<<" "<<z<<" "<<w<<endl;
using namespace std;
//using Modint = atcoder::modint998244353;
typedef long long ll;
void ex_gcd(ll a,ll b, ll& d,ll& x,ll& y){
  if(!b) {d=a;x=1;y=0;}
  else   {	ex_gcd(b,a%b,d,y,x);	y-=a/b*x;}
}
ll inv(ll a,ll n)  {
  ll d,x,y;	ex_gcd(a,n,d,x,y);
  return d==1?(x%n+n)%(n/d):-1;
}
ll gcd(ll x, ll y){
  if(y==0)return x;
  return gcd(y, x%y);
}

const int maxm = 50005;
const int maxn = 25;
const int mod = 998244353;

int a[maxn][maxm];

int main() {
#ifdef suiyuan2009
  freopen("/Users/suiyuan2009/CLionProjects/icpc/input.txt", "r", stdin);
  //freopen("/Users/suiyuan2009/CLionProjects/icpc/output.txt", "w", stdout);
#endif
  int n,m;
  cin>>n>>m;
  for(int i=0;i<n;i++){
    for(int j=0;j<m;j++){
      cin>>a[i][j];
    }
  }
  ll all = 1;
  for(int i=2;i<=n;i++)all*=i;
  ll sum = 0;
  for(int i=0;i<m;i++){
    vector<int>v;
    ll cur = 1;
    for(int j=0;j<n;j++){
      v.push_back(a[j][i]-1);
    }
    sort(v.begin(),v.end());
    for(int j=0;j<n;j++){
      int tt=v[j]-j;
      if(tt<=0)cur = 0;
      else cur*=tt;
    }
    cur = all -cur;
    sum=(sum+cur%mod)%mod;
  }
  ll ret = sum*inv(all%mod, mod)%mod;
  cout<<ret<<endl;
  return 0;
}