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}
/*
西江月·证明
即得易见平凡，仿照上例显然；留作习题答案略，读者自证不难。
反之亦然同理，推论自然成立；略去过程Q.E.D.，由上可知证毕。

start thinking:
BULB:
result of thinking:

start coding:
AC:
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldouble;
template<class T> bool chmin(T &x, const T &y) {
  return x > y ? (x = y, true) : false;
}
template<class T> bool chmax(T &x, const T &y) {
  return x < y ? (x = y, true) : false;
}

#define maxn 25
#define maxm 50005
const ll mod = 998244353;
int n, m, d[maxn][maxm];
ll fct[maxn];

ll modpow(ll x, ll y) {
  ll res = 1;
  while (y) {
    if (y & 1)
      res = res * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return res;
}
ll inv(ll x) {
  return modpow(x, mod - 2);
}

int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%d", &d[i][j]);
    }
  }
  fct[0] = 1;
  for (int i = 1; i <= n; i++)
    fct[i] = fct[i - 1] * i % mod;
  ll ans = 0;
  for (int i = 1; i <= m; i++) {
    vector<int> ds;
    for (int j = 1; j <= n; j++)
      ds.push_back(d[j][i]);
    sort(ds.begin(), ds.end());
    ll bad = 1;
    for (int j = 0; j < n; j++) {
      if (ds[j] <= j + 1)
        bad = 0;
      else
        (bad *= ds[j] - 1 - j) %= mod;
    }
    (ans += fct[n] - bad) %= mod;
  }
  (ans *= inv(fct[n])) %= mod;
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}
