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 pb push_back   
#define mp make_pair     
#define nline "\n"           
#define debug(x) cerr<<#x<<" = "<<x<<" "   
#define dnl cerr<<"\n"                       
const ll N=2e12+5;     
const ll M=998244353;  
ll binpow(ll a,ll b){
    if(b==0)
        return 1;
    ll x=binpow(a,b/2);
    x=(x*x)%M;
    if(b%2)
        return (x*a)%M;
    return x;
}  
void solve(){  
    ll n,m; cin>>n>>m;
    vector<ll> dist[m+2];  
    ll fact[n+2];    
    fact[0]=1;  
    for(ll i=1;i<=n;i++)
        fact[i]=(fact[i-1]*i)%M;
    for(ll i=1;i<=n;i++){
        for(ll j=1;j<=m;j++){
            ll x; cin>>x;  
            dist[j].pb(x);
        }
    }
    ll conquered=0;
    for(ll i=1;i<=m;i++){
        ll now=1;
        vector<ll> smaller(n+3,0);
        for(ll x:dist[i]){
            for(ll j=1;j<x;j++)
                smaller[j]++;
        }
        ll sub=0;
        for(ll i=n;i>=1;i--){
            ll mul=smaller[i]-sub;
            if(mul<=0)  
                now=0;
            else
                now=(now*mul)%M;
            sub++;
        }
        conquered=((conquered+fact[n]-now)%M+M)%M;
    }
    ll inv=binpow(fact[n],M-2);
    conquered=(conquered*inv)%M;
    cout<<conquered;
}
int main()                               
{         
    ios_base::sync_with_stdio(false);                      
    cin.tie(NULL);                     
    #ifndef ONLINE_JUDGE       
    freopen("input.txt", "r", stdin);                                     
    freopen("output.txt", "w", stdout);  
    freopen("error.txt", "w", stderr);                        
    #endif             
    ll t=1;
    //cin>>t;  
    while(t--){
        solve();  
    }
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}                                