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>

typedef long long ll;
typedef long double ld;
#define FASTIO ios_base::sync_with_stdio(false); cin.tie(NULL);
#define PRECISION cout << fixed << setprecision(20);

using namespace std;

const int mod = 998244353;

struct Mint {
    ll v; explicit operator ll() const { return v; }
    Mint() { v = 0; }
    Mint(ll _v) { 
        v = (-mod < _v && _v < mod) ? _v : _v % mod;
        if (v < 0) v += mod;
    }
    friend bool operator==(const Mint& a, const Mint& b) { return a.v == b.v; }
    friend bool operator!=(const Mint& a, const Mint& b) { return !(a == b); }
    friend bool operator<(const Mint& a, const Mint& b) { return a.v < b.v; }
   
    Mint& operator+=(const Mint& m) { if ((v += m.v) >= mod) v -= mod; return *this; }
    Mint& operator-=(const Mint& m) { if ((v -= m.v) < 0) v += mod; return *this; }
    Mint& operator*=(const Mint& m) { v = v*m.v%mod; return *this; }
    Mint& operator/=(const Mint& m) { return (*this) *= inv(m); }
    Mint& operator^=(const ll& e) { return (*this) = pow(*this, e); }

    friend Mint pow(Mint a, ll p) {
        Mint ans = 1; assert(p >= 0);
        for (; p; p /= 2, a *= a) if (p & 1) ans *= a;
        return ans;
    }
    
    friend Mint inv(const Mint& a) { assert(a.v != 0); return pow(a, mod-2); }

    Mint operator-() const { return Mint(-v); }
    Mint& operator++() { return *this += 1; }
    Mint& operator--() { return *this -= 1; }
    Mint operator++(int) { Mint temp; temp.v = v++; return temp; }
    Mint operator--(int) { Mint temp; temp.v = v--; return temp; }

    friend Mint operator+(Mint a, const Mint& b) { return a += b; }
    friend Mint operator-(Mint a, const Mint& b) { return a -= b; }
    friend Mint operator*(Mint a, const Mint& b) { return a *= b; }
    friend Mint operator/(Mint a, const Mint& b) { return a /= b; }
    friend Mint operator^(Mint a, const ll& e) { return a ^= e; }

    friend ostream& operator<<(ostream& os, const Mint& m) {
        os << m.v; return os;
    }

    friend istream& operator>>(istream& is, Mint& m) {
        ll x; is >> x;
        m.v = x;
        return is;
    }
};

void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>> d(n, vector<int>(m));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> d[i][j];
        }
    }

    Mint f = 1;

    Mint ans = 0;
    for (int j = 0; j < m; j++) {
        vector<int> store;
        for (int i = 0; i < n; i++) {
            store.push_back(d[i][j] - 1);
        }
        int sub = 0;
        Mint curr = 1;
        sort(store.begin(), store.end());
        int rem = 0;
        for (int s: store) {
            int have = s - rem;
            if (have <= 0) {
                curr = 0; break;
            } else {
                curr *= have;
            }
            rem++;
        }
        ans += curr;
    }

    for (int i = 1; i <= n; i++) f *= i;

    cout << (m * f - ans) / f << "\n";
}

int main() {
    FASTIO;
    PRECISION;

    int t = 1;
    // cin >> t;

    for (int i = 1; i <= t; i++) {
        solve();
    }
}