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;

const int64_t MOD = 998244353;

int64_t Factorial(const int64_t& x) {
    if (x == 0) { return 1LL; }
    int64_t fact = 1LL;
    for(int64_t i=1; i<=x; i++) {
        fact *= i;
    }
    return fact;
}

int64_t binpow (int64_t a, int64_t n, int64_t m) {
	if (n == 0)
		return 1LL;
	if (n % 2 == 1)
		return ((binpow(a, n-1, m) % m) * (a % m)) % m;
	else {
		int64_t b = binpow (a, n/2, m) % m;
		return (b*b) % m;
	}
}


int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    int n, m;
    cin >> n >> m;

    vector<vector<int>> r(n, vector<int>(m));
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            cin >> r[i][j];
            --r[i][j];
        }
    }

    int64_t n_fact = Factorial(n);
    int64_t y_rev = (binpow(n_fact, MOD-2, MOD))%MOD;

    int64_t final_expectation = 0LL;
    // Using linearity of expectation
    // find it for every single point
    for(int j=0; j<m; j++) {
        vector<int> dist(n+1, 0);
        for(int i=0; i<n; i++) {
            dist[ r[i][j] ]++;
        }

        int64_t uncover_perm = 1LL;
        int places_filled = 0;
        int used_cities = 0;
        for(int k=0; k<=n; k++) {
            used_cities += dist[k];
            if(used_cities > k) {
                uncover_perm = 0LL;
                break;
            }

            int pl_k = k - places_filled;
            uncover_perm *= Factorial(pl_k);
            uncover_perm /= Factorial(pl_k - dist[k]);
            places_filled += dist[k];
        }

        int64_t cover_perm = n_fact - uncover_perm;
        final_expectation += (cover_perm % MOD);
    }

    final_expectation %= MOD;
    final_expectation *= y_rev;

    cout << final_expectation % MOD;
    return 0;
}
