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 pb push_back
#define ff first
#define ss second
#define S pair<pair<int, int>, pair<int, int>>
#define n4 ff.ff
#define n7 ff.ss
#define n74 ss.ff
#define n47 ss.ss


typedef long long ll;
typedef pair<ll, ll> pii;

const int N = 1e5+8;
const int bits = 30;
const ll mod = 998244353;
int n, h, m;

int puntos[20][N];

ll sum(ll a, ll b) {
    ll rta = a+b;
    return rta>= mod ? rta-mod : rta;
}

ll sub(ll a, ll b) {
    ll rta = a-b+mod;
    return rta>= mod ? rta-mod : rta;
}

ll multi(ll a, ll b) {
    ll rta = a*b;
    return rta>= mod ? rta%mod : rta;
}

ll inv (ll x) {
    ll rta = 1;
    int e = mod-2;
    while(e) {
        if(e&1) {
            rta = multi(rta, x);
        }
        x = multi(x, x);
        e>>=1;
    }
    return rta;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);


    int nn, T,  a, b, h, x, m;
    char c;
    T = 1;

    while(T--) {
        cin>>n>>m;
        for(int i = 0; i<n; ++i) {
            for(int j = 0; j<m; ++j) {
                cin>> puntos[i][j];
            }
        }

        ll num = 0;
        ll nf = 1, nfi;
        for(int i = 1; i<=n; ++i) nf = multi(nf, i);
        nfi = inv(nf);
        for(int i = 0; i<m; ++i) {
            int cnt[n+1];
            ll acum = 0, mul = 1;
            for(int j = 0; j<n+1; ++j) cnt[j] = 0;

            for(int j = 0; j<n; ++j) cnt[n+1-puntos[j][i]]++;

            for(int j = 0; j<n; ++j) {
                acum+=cnt[j];
                mul= multi(mul, acum-j);
            }
            if(cnt[n]) mul = 0;
            num = sum(num, sub(nf, mul));
        }
        num = multi(num, nfi);
        cout<<num<<"\n";
    }
    return 0;
}
