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 <algorithm>
#include <assert.h>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
using namespace std;

#define ll long long
#define ull unsigned long long
#define MOD 998244353LL
#define INFD INT_MAX
#define INFL 1000000000000000000LL
#define eps 1e-9
#define PI 3.14159265358979
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define pow __pow
#define gcd __gcd
#define count __count
#define sd(x) scanf("%d", &(x))
#define slld(x) scanf("%lld", &(x))
#define pd(x) printf("%d\n", (x))
#define plld(x) printf("%lld\n", (x))
#define gc getchar
#define pll pair<ll, ll> 
#define contains(S, x) (S.find(x) != S.end())
#define loop(it, st, en) for(ll it = (st); it <= (en); it++)

ll pow(ll a, ll e) {
	if(e == 0) {
		return 1;
	}
	ll h = pow(a, e / 2);
	h = (h * h) % MOD;
	if(e & 1) {
		h = (h * a) % MOD;
	}
	return h;
}

const ll N = 20;
const ll D = 20;
const ll M = 5e4;

ll arr[M][N];
ll dp[D + 1];

int main() {
	// freopen("in", "r", stdin);
	// freopen("out", "w", stdout);
	ll n, m; cin >> n >> m;
	for(ll i = 0; i < n; i++) {
		for(ll j = 0; j < m; j++) {
			cin >> arr[j][i];
		}
	}
	ll totalUndesirables = 0;
	for(ll j = 0; j < m; j++) {
		for(ll d = 0; d <= n; d++) {
			dp[d] = 0;
		}
		for(ll i = 0; i < n; i++) {
			ll a = arr[j][i];
			dp[a - 1]++;
		}
		for(ll d = n - 1; d >= 0; d--) {
			dp[d] += dp[d + 1];
		}
		ll undesirables = 1;
		for(ll d = n; d >= 1; d--) {
			undesirables = (undesirables * (dp[d] - (n - d))) % MOD;
		}
		totalUndesirables += undesirables;
	}
	totalUndesirables %= MOD;
	ll invFacN = 1;
	for(ll i = 2; i <= n; i++) {
		invFacN = (invFacN * pow(i, MOD - 2)) % MOD;
	}
	ll ans = (m - ((invFacN * totalUndesirables) % MOD) + MOD) % MOD;
	cout << ans << endl;
}