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>
using namespace std;
 
const int N = 25;
const int M = 50005;
const int MOD = 998244353;

int n, m;
int d[N][M];
int fact[N];
int inv[N];

int power(int x, int y) {
  int ret = 1;
  for (; y; y >>= 1) {
    if (y & 1) ret = 1LL * ret * x % MOD;
    x = 1LL * x * x % MOD;
  }
  return ret;
}

void init() {
  fact[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = 1LL * fact[i - 1] * i % MOD;
    inv[i] = power(i, MOD - 2);
  }
}

int compute(vector<int> v) {
  sort(v.begin(), v.end(), greater<int>());
  int ret = 0;
  int cur = 1;
  for (int i = 0; i < v.size(); ++i) {
    ret = (1LL * cur * min(v[i], n - i) % MOD * inv[n - i] + ret) % MOD;
    cur = 1LL * cur * max(0, (n - i) - v[i]) % MOD * inv[n - i] % MOD;
    // cerr << v[i] << " " << ret << " " << cur << endl;
  }
  // cerr << ret << endl;
  return ret;
}

int solve() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      scanf("%d", &d[i][j]);
    }
  }

  int ans = 0;
  for (int j = 0; j < m; ++j) {
    vector<int> v(n);
    for (int i = 0; i < n; ++i) {
      v[i] = (n + 1) - d[i][j];
    }
    ans = (ans + compute(v)) % MOD;
  }

  printf("%d\n", ans);
  return 0;
}

int main() {
  init();
  int t = 1;
  // scanf("%d", &t);
  for (int tc = 0; tc < t; ++tc) {
    // printf("Case #%d: ", tc+1);
    solve();
  }
  return 0;
}
