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}
// Problem: E. Assimilation IV
// Contest: Codeforces - Educational Codeforces Round 109 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1525/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;

#define SPEED ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long
#define endl '\n'
#define vi vector<int>
#define pii pair<int, int>
#define pb push_back
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define fill(a,b) memset(a, b, sizeof(a))
#define setbits(x) __builtin_popcountll(x)

const int inf=0x3f3f3f3f3f3f3f3f, mod = 998244353;
int dx[]={0,1,1,1,0,-1,-1,-1};
int dy[]={1,1,0,-1,-1,-1,0,1};

// <--------------------Code Begins Here-------------------->
int powmod(int a, int b, int p){
	int res = 1;
	while(b != 0) {
		if(b & 1)
		  res = (res % p * a % p) % p;
		a = (a%p * a%p)%p;
		b /= 2;
	}
	return res%p;
}
int32_t main() {
    SPEED;
    int t = 1;
    // cin>>t;
    while(t--) {
    	int n, m;
    	cin>>n>>m;
    	int d[n][m];
    	for(int i = 0; i < n; ++i) {
    		for(int j = 0; j < m; ++j) {
    			cin>>d[i][j];
    		}
    	}
    	int fac = 1;
    	for(int i = 2; i <= n; ++i) {
    		fac *= i;
    		fac %= mod;
    	}
    	int ans = 0;
    	for(int j = 0; j < m; ++j) {
    		map<int, int> mp;
    		int mn = n + 1;
    		for(int i = 0; i < n; ++i) {
    			mn = min(mn, d[i][j] - 1);
    			for(int k = 1; k < d[i][j]; ++k) {
    				mp[k]++;
    			}
    		}
    		if(mn == 0) {
    			ans += fac;
    			ans %= mod;
    			continue;
    		}
    		int val = 1, cur = 0;
    		for(int i = n; i >= 1; --i) {
    			if(mp[i] - cur <= 0) {
    				val = 0;
    				break;
    			}
    			val *= (mp[i] - cur);
    			val %= mod;
    			++cur;
    		}
    		ans += (fac - val);
    		ans %= mod;
    		if(ans < 0)ans += mod;
    	}
    	ans *= powmod(fac, mod - 2, mod);
    	ans %= mod;
    	cout<<ans;
    }
    return 0;
}