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}
/*#pragma GCC optimize("Ofast")
 #pragma GCC optimize ("unroll-loops")
 #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 #pragma comment(linker, "/stack:200000000")*/
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
#include <stack>
#include <deque>
#include <iterator>
#include <tuple>
#include <iomanip>
#include <valarray>
#include <limits>
#include <sstream>
#include <cassert>
#include <functional>
#include <numeric>
#define in insert
#define pb push_back
#define mp make_pair
#define int long long
#define fi first
#define double long double
#define se second
#define endl '\n'
#define pii pair<int,int>
#define pdd pair<double,double>
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define mod 998244353
#define MAXN  101
#define all(v) v.begin(),v.end()
using namespace std;
/*using namespace __gnu_pbds;
 typedef tree<int, null_type, less<int>, rb_tree_tag,
 tree_order_statistics_node_update>
 ordered_set;*/
int spf[MAXN];
int fac[MAXN];
void sieve()//sieve
{
    spf[1] = 1;
    for (int i=2; i<MAXN; i++)
        spf[i] = i;
    for (int i=4; i<MAXN; i+=2)
        spf[i] = 2;
    for (int i=3; i*i<MAXN; i++)
    {
        if (spf[i]==i)
        {
            for (int j=i*i; j<MAXN; j+=i)
                if (spf[j]==j)
                    spf[j]=i;
        }
    }
}//sieve ends
map<int,int> getfactor(int a)//factor prime
{
    map<int,int> m;
    while(a>1)
    {
        m[spf[a]]++;
        a/=spf[a];
    }
    return m;
}//prime factor ends
int power(int x,int y,int p)
{
    int res = 1;
    x = x%p;
    if (x == 0||y < 0) return 0;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x)% p;
        y = y>>1;
        x = (x*x)%p;
    }
    return res;
}
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
int inverse(int a, int p)
{
    return power(a, p-2, p);
}
int ncr(int n, int r, int p)
{
    if(r>n)
        return 0;
    if (r==0||r==n)
        return 1;
    return (fac[n]*inverse(fac[r], p) % p *
            inverse(fac[n-r], p) % p) % p;
}
void solve()
{
    int n,m;
    cin>>n>>m;
    int a[n+1][m+1];
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            cin>>a[i][j];
    int ans=(m*(fac[n]))%mod;
    for(int j=1;j<=m;j++)
    {
        int leftover=0;
        int filled=0;
        map<int,int>map;
        for(int i=1;i<=n;i++)
        {
            if(a[i][j]>n)
                leftover++;
            else map[a[i][j]]++;
        }
        int temp=1;
        for(auto it:map)
        {
            temp=temp*(ncr(it.fi-filled-1,it.se,mod)*fac[it.se])%mod;
            temp%=mod;
            filled+=it.se;
        }
        temp*=fac[leftover];
        temp%=mod;
        ans=(ans-temp+mod)%mod;
    }
    cout<<(ans*inverse(fac[n],mod))%mod<<endl;
}
signed main()
{
    fast
    int t=1;
    //cin>>t;
    fac[0]=1;
    for(int i=1;i<=100;i++)
        fac[i]=(fac[i-1]*i)%mod;
    while(t--)
    {
        solve();
    }
}
