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}
/*
    author:  Maksim1744
    created: 16.05.2021 11:21:34
*/

#include "bits/stdc++.h"

using namespace std;

using ll = long long;
using ld = long double;

#define mp   make_pair
#define pb   push_back
#define eb   emplace_back

#define sum(a)     ( accumulate ((a).begin(), (a).end(), 0ll))
#define mine(a)    (*min_element((a).begin(), (a).end()))
#define maxe(a)    (*max_element((a).begin(), (a).end()))
#define mini(a)    ( min_element((a).begin(), (a).end()) - (a).begin())
#define maxi(a)    ( max_element((a).begin(), (a).end()) - (a).begin())
#define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())

template<typename T>             vector<T>& operator--            (vector<T> &v){for (auto& i : v) --i;            return  v;}
template<typename T>             vector<T>& operator++            (vector<T> &v){for (auto& i : v) ++i;            return  v;}
template<typename T>             istream& operator>>(istream& is,  vector<T> &v){for (auto& i : v) is >> i;        return is;}
template<typename T>             ostream& operator<<(ostream& os,  vector<T>  v){for (auto& i : v) os << i << ' '; return os;}
template<typename T, typename U> pair<T,U>& operator--           (pair<T, U> &p){--p.first; --p.second;            return  p;}
template<typename T, typename U> pair<T,U>& operator++           (pair<T, U> &p){++p.first; ++p.second;            return  p;}
template<typename T, typename U> istream& operator>>(istream& is, pair<T, U> &p){is >> p.first >> p.second;        return is;}
template<typename T, typename U> ostream& operator<<(ostream& os, pair<T, U>  p){os << p.first << ' ' << p.second; return os;}
template<typename T, typename U> pair<T,U> operator-(pair<T,U> a, pair<T,U> b){return mp(a.first-b.first, a.second-b.second);}
template<typename T, typename U> pair<T,U> operator+(pair<T,U> a, pair<T,U> b){return mp(a.first+b.first, a.second+b.second);}
template<typename T, typename U> void umin(T& a, U b){if (a > b) a = b;}
template<typename T, typename U> void umax(T& a, U b){if (a < b) a = b;}

#ifdef HOME
#define SHOW_COLORS
#include "C:/C++ libs/print.cpp"
#else
#define show(...) void(0)
#define mclock    void(0)
#define shows     void(0)
#define debug  if (false)
#endif

#define ALL(x) x.begin(), x.end()


struct mod_int {
    ll x;
    static const ll MD = 998244353;
    mod_int(ll p = 0) {x = p;}
    friend mod_int operator + (const mod_int &a, const mod_int &b) { return (a.x + b.x) % MD;}
    friend mod_int operator - (const mod_int &a, const mod_int &b) { return (MD + a.x - b.x) % MD;}
    friend mod_int operator * (const mod_int &a, const mod_int &b) { return (a.x * b.x) % MD;}
    friend mod_int pow(mod_int a, ll n) {
        if (n == 1) return a;
        return pow(a*a, n / 2) * (n&1? a: 1);
    }
    friend mod_int operator / (const mod_int &a, const mod_int &b) { return a * pow(b, MD-2);}
    friend ostream &operator << (ostream &o, const mod_int &a) { return o << a.x;}
    friend istream &operator >> (istream &i, mod_int &a) { return i >> a.x;}
};

int main() {
    int n, m; cin >> n >> m;
    vector<mod_int> fact(n+1);
    fact[0] = 1;
    for (int i = 1; i <= n; i++) fact[i] = fact[i-1] * i;
    mod_int gd = 0;
    vector<vector<int>> dt(m, vector<int>(n));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> dt[j][i];
        }
    }
    for (int i = 0; i < m; i++) {
        sort(ALL(dt[i]));
        vector<int> q(n, n);
        for (int j = 0; j < n; j++) 
            for (int k = dt[i][j] - 1; k < n; k++)
                q[k]--;
        mod_int bd = 1;
        for (int j = n - 1; j >=0 ; j--) 
            bd = bd * (q[j] - n + j + 1);
        gd = gd + fact.back() - bd;
    }
    cout << gd / fact.back();
    return 0;
}