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>
typedef long long ll;
#define vll vector<ll>
#define pb push_back
#define pll pair<ll,ll>
#define vpll vector<pair<ll,ll> >
#define st(x) sort(x.begin(),x.end())
#define sz(x) (ll)x.size()
#define input(n,x) for(ll i=0;i<n;i++){ll k;cin>>k;x.pb(k);}
#define fr first
#define sc second
#define rep(i,j,k) for(ll i=j;i<k;i++)
#define repn(i,j,k) for(ll i=j;i>=k;i--)
using namespace std;
vll x;
const ll mod =  998244353;
ll inv(ll a, ll b) {
	if (b == 0) {
		return 1;
	}
	if (b == 1) {
		return a;
	}
	ll k = inv(a, b / 2);
	k = (k * k) % mod;
	if (b & 1) {
		return (k * a) % mod;
	}
	return k;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	// IF USE OF LEAST / MIN  ANYWHERE IN THE STATEMENT FIRST THOUGHT SHOULD BE "BINARY SEARCH"
	// THINK DP WHENEVER N*M<=1000000 AT MAX
	// WRITE DOWN EVERYTHING
	// FOR FLOWS USE DINICS ALGORITHM , TEMPLATE IS PRESENT IN ROCK PAPER SCISSOR PROBLEM
	// CHECK FOR GCD , LCM APPROACH . HARD TO SPOT
	// for problems related to subsequence the best dp is - the number of  subsequences that end at index i and have length j is equal to addition of the number of subsequences that end at index (0 ... i-1) and hve length j-1

	ll n, m;
	cin >> n >> m;
	ll b[n + 1][m + 1];
	rep(j, 0, n) {
		rep(k, 0, m) {
			cin >> b[j][k];
		}
	}
	ll f = 1;
	rep(j, 2, n + 1) {
		f = (f * j) % mod;
	}
	ll ans = 0;
	rep(j, 0, m) {
		vll pos;
		rep(k, 0, n) {
			pos.pb(min(n, b[k][j] - 1));
		}
		st(pos);
		ll t = 1;
		rep(j, 0, n) {
			t = (t * max(ll(0), pos[j] - j)) % mod;
		}
		ans = (ans + (f - t + mod) % mod + mod) % mod;
	}
	cout << (ans * inv(f, mod - 2)) % mod << "\n";
	return 0;
}