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;

constexpr int mod = 998244353;

int dp[23][23];

int cnk[23][23];

void init() {
  for (int i = 0; i <= 20; i++) {
    cnk[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      cnk[i][j] = cnk[i - 1][j] + cnk[i - 1][j - 1];
    }
  }
}

int mul(int x) {
  return x;
}

template <typename... Args>
int mul(int x, Args... args) {
  return 1ll * x * mul(args...) % mod;
}

int calculate(int n, std::vector<int>& cnt) {
  for (int i = 1; i <= n + 1; i++) {
    for (int j = 0; j <= n; j++) {
      dp[i][j] = 0;
    }
  }
  dp[1][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (dp[i][j] == 0) {
        continue;
      }
      int x = 1;
      for (int k = 0; k <= cnt[i]; k++) {
        dp[i + 1][j + k] += mul(dp[i][j], cnk[cnt[i]][k], x);
        if (dp[i + 1][j + k] >= mod) {
          dp[i + 1][j + k] -= mod;
        }
        x = mul(x, i - (j + k));
      }
    }
  }
  int ans = 0;
  for (int j = 1; j <= n; j++) {
    int x = 1;
    for (int k = n - j; k >= 1; k--) {
      x = mul(x, k);
    }
    if (j % 2 == 1) {
      ans += mul(x, dp[n + 1][j]);
      if (ans >= mod) {
        ans -= mod;
      }
    } else {
      ans += mod - mul(dp[n + 1][j], x);
      if (ans >= mod) {
        ans -= mod;
      }
    }
  }
  return ans;
}

int pw(int n, int k) {
  if (k == 0) {
    return 1;
  }
  int g = pw(n, k / 2);
  g = mul(g, g);
  if (k % 2 == 1) {
    g = mul(g, n);
  }
  return g;
}

int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  std::cin >> n >> m;
  init();
  std::vector<std::vector<int>> d(n);
  for (int i = 0; i < n; i++) {
    d[i].resize(m);
    for (int j = 0; j < m; j++) {
      cin >> d[i][j];
    }
  }
  int ans = 0;
  for (int j = 0; j < m; j++) {
    std::vector<int> cnt(n + 1);
    for (int i = 0; i < n; i++) {
      if (d[i][j] <= n) {
        cnt[n - d[i][j] + 1]++;
      }
    }
    ans += calculate(n, cnt);
    if (ans >= mod) {
      ans -= mod;
    }
  }
  int cnt = 1;
  for (int i = 1; i <= n; i++) {
    cnt = mul(cnt, i);
  }
  cerr << ans << "\n";
  cout << mul(ans, pw(cnt, mod - 2)) << "\n";
  return 0;
}
