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>
#define ull unsigned long long
#define ll long long
#define il inline
#define db double
#define ls rt << 1
#define rs rt << 1 | 1
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define X first
#define Y second
#define pcc pair<char, char>
#define vi vector<int>
#define vl vector<ll>
#define rep(i, x, y) for(int i = x; i <= y; i ++)
#define rrep(i, x, y) for(int i = x; i >= y; i --)
#define rep0(i, n) for(int i = 0; i < (n); i ++)
#define per0(i, n) for(int i = (n) - 1; i >= 0; i --)
#define ept 1e-9
#define sz(x) (x).size()
#define All(x) (x).begin(), (x).end()
using namespace std;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
inline ll read1()
{
	ll x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
#define int long long
const int N = 50010;
const int mod = 998244353;
int n, m;
int d[N][25];
void add(int &a, int b) {a += b; if (a >= mod) a -= mod;}
void dec(int &a, int b) {a -= b; if (a < 0) a += mod;}
int pow_mod(int a, int i)
{
	int s = 1;
	while (i)
	{
		if (i & 1) s = 1LL * s * a % mod;
		a = 1LL * a * a % mod;
		i >>= 1;
	}
	return s;
}
signed main()
{
	n = read();
	m = read();
	rep0(i, n)
		rep0(j, m)
			d[j][i] = read();
	rep0(i, m) sort(d[i], d[i] + n);
	int ans = m;
	int jc = 1;
	rep(i, 1, n) jc = 1LL * jc * i % mod;
	jc = pow_mod(jc, mod - 2);
	rep0(i, m)
	{
		int res = 1;
		rrep(j, n - 1, 0) res = 1LL * res * max(0LL, d[i][j] - (j + 1)) % mod;
		dec(ans, 1LL * res * jc % mod);
	}
	printf("%d\n", ans);
	return 0;
}
