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;
typedef long long ll;

ll a[25][50005];

ll f[25];
ll finv[25];
ll M = 998244353;
ll cnt[25];

ll fp(ll x,ll k){
    ll r = 1;
    while(k){
        if(k&1){
            r = r * x % M;
        }
        x = x * x % M;
        k>>=1;
    }
    return r;
}

int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n,m;
cin>>n>>m;
for(int i=1;i<=n;++i){
    for(int j=1;j<=m;++j){
        cin>>a[i][j];
    }
}
f[0] = 1;
for(int i=1;i<=n;++i){
    f[i] = f[i-1] * i % M;
}
finv[n] = fp(f[n],M-2);
for(int i=n-1;i>=0;--i){
    finv[i] = finv[i+1] * (i+1) % M;
}

ll total = 0;
for(int i=1;i<=m;++i){
    ll remcount = 1;
    ll count = 0;
    for(int j=0;j<=21;++j){
        cnt[j] = 0;
    }
    for(int j=1;j<=n;++j){
        cnt[a[j][i]] ++;
    }
    for(int j=2;j<=n;++j){
        cnt[j] += cnt[j-1];
    }
    for(int j=n;j>=1;--j){
        count += remcount * cnt[j] % M * f[j-1] % M;
        remcount = remcount * (j-cnt[j]) % M;
            //cout<<j<<": "<<count<<" "<<remcount<<endl;
    }
    //cout<<count<<endl;
    total = (total + count) % M;
}
cout<<(total * finv[n] % M + M) % M <<endl;

}