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;

using ll = long long;
using ld = long double;

int a[25][50005];

const int MOD = 998244353;

int add(int a, int b){
    a += b;
    if(a >= MOD) a -= MOD;
    return a;
}

int mul(int a, int b){
    return (1LL*a*b)%MOD;

}

int pw(int a, int b){
    if(b == 0) return 1;
    int res = pw(a, b/2);
    res = mul(res, res);
    if(b%2) res = mul(res, a);
    return res;
}

int main(){
    ios_base::sync_with_stdio(false), cin.tie(0);
    cout.precision(10);
    cout << fixed;

    int n, m;
    cin >> n >> m;
    for(int i=1; i<=n; i++){
        for(int j=1; j<=m; j++){
            cin >> a[i][j];
            a[i][j]--;
        }
    }
    int res = 0;
    for(int j=1; j<=m; j++){
        vector <int> vec;
        for(int i=1; i<=n; i++){
            vec.push_back(a[i][j]);
        }
        sort(vec.begin(), vec.end());
        int ways = 1;
        for(int g=0; g<vec.size(); g++){
            ways = mul(ways, vec[g] - g);
        }
        if(ways == 0) continue;
        res = add(res, ways);
    }
    int fact = 1;
    for(int i=1; i<=n; i++) fact = mul(fact, i);
    int p = add(mul(fact, m), MOD - res);
    p = mul(p, pw(fact, MOD-2));
    cout << p;
    return 0;
}
