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}
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <iomanip>
#include <bitset>
#include <unordered_map>
#include <stack>
#include <memory.h>
#include <list>
#include <numeric>
#include <functional>
#include <complex>
#include <cassert>
#include <regex>
#include <random>
#include <sstream>

#define ll long long
#define ld long double
#define ull unsigned ll
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define pii pair<int,int> 
#define vi vector<int>
#define vll vector<ll>
#define vpii vector<pii>
#define vvi vector<vector<int>>
#define forn(it,from,to) for(int (it)=from; (it)<to; (it)++)
ll LINF = (ll)4e18;
ll mod = 998244353;
ll mod2 = 1e9 + 7;
using namespace std;
const int INF = 1000 * 1000 * 1000;

ll bp(ll x, ll y) {
	if (y == 0) return 1;
	if (y % 2) return x * bp(x * x % mod, y / 2) % mod;
	else return bp(x * x % mod, y / 2);
}

ll inv(ll x) {
	return bp(x, mod - 2);
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#else
	//freopen("sum.in", "r", stdin);
	//freopen("sum.out", "w", stdout);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
#endif
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	cout.setf(ios::fixed);
	int n, m;
	cin >> n >> m;
	vvi a(n, vi(m));
	ll fact[21];
	fact[0] = 1;
	forn(i, 1, 21) {
		fact[i] = fact[i - 1] * i;
		fact[i] %= mod;
	}
	forn(i, 0, n) {
		forn(j, 0, m) cin >> a[i][j];
	}
	ll ans = 0;
	forn(i, 0, m) {
		vi tmp;
		ll cur = 1;
		forn(j, 0, n) {
			if (n - a[j][i] + 1 >= n) {
				cur = 0;
				break;
			}
			tmp.push_back(n - (n - a[j][i] + 1) - 1);
		}
		if (cur != 0) {
			sort(all(tmp));
			int d = n - tmp.size();
			ll cnt = 0;
			for (auto& t : tmp) {
				cur *= t + 1 - cnt;
				cnt++;
				cur %= mod;
			}
			if (cnt > 0) {
				cur *= fact[d];
				cur %= mod;
			}
			else cur = 0;
		}
		ll kek = fact[n] - cur;
		kek %= mod;
		kek += mod;
		kek %= mod;
		ans += kek;
		ans %= mod;
	}

	ll zn = inv(fact[n]);
	ans *= zn;
	ans %= mod;

	cout << ans;


	return 0;
}