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: vamp-coder
// Practise makes a man perfect
// Never give up

#include <bits/stdc++.h>


using namespace std;

#define ll long long 
#define int ll
#define pb push_back
#define vi vector<int>
#define vl vector<ll>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define enter cout<<"\n"


// -------------- Arithmetic --------------
const ll MOD = 998244353;

ll power(ll x, ll y, ll m) {
    ll ans = 1;
    x = x % m;
    if (x == 0) return 0;
    while (y > 0) {
        if (y & 1) ans = (ans * x) % m;
        y = y >> 1;
        x = (x * x) % m;
    }
    return ans;
}

ll mod_inv(ll a, ll m) {return power(a, m - 2, m);}
ll mod_(ll a, ll m) {return (a+m)%m;}
ll add(ll a, ll b, ll m) {return (((a + b) % m) + m) % m;}
ll sub(ll a, ll b, ll m) {return (((a - b) % m) + m) % m;}
ll mul(ll a, ll b, ll m) {return (((a * b) % m) + m) % m;}
ll div_(ll a, ll b, ll m) {return mul(a, mod_inv(b, m), m);}

// -------- Factorial -----------
const ll N1 = 30;
ll f[N1],invf[N1];
void fact(ll M){
    f[0]=f[1]=1;
    for(int i=2;i<N1;i++)
        f[i]=mod_(f[i-1]*i, M);
 
    invf[N1-1]=power(f[N1-1],M-2, M);
    for(int i=N1-1;i>0;i--)
        invf[i-1]=mod_(invf[i]*i, M);
 
}
 
// function to calculare nCr mod M
ll nCr(ll n,ll r, ll M) {
    if(r>n)
        return 0;
    ll ans = f[n];
    ans = mod_(ans*invf[n-r], M);
    ans = mod_(ans*invf[r], M);
    return ans;
}

/*
    IMPORTANT: if (x%a)%b == (x%b)%a, and a<=b then
    x = K*(LCM(a,b)) + r', where 0 <= r' < b, and K is some integer.

    If we want to find such x in range [0,R], 
    then the x which satisfies the above property also satisfies:

    1. K belongs to [0, floor(R / LCM(a,b)) - 1] and r' belongs to [0,b)
    (for all K and r in above point, the resulting x will be in answer)

    2. When K = floor(R / LCM(a,b)), 
    then we will take only those r' such that resultant x <= R.

*/


// ---------- Fast IO -----------
void fastIO() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    cout << std::fixed << setprecision(25);
    cerr << std::fixed << setprecision(10);
}

// -------- some helper functions --------

string reverse(string s) {
    int n = s.size();
    string ans = "";
    for(int i =n-1; i>=0; i--) {
        ans.push_back(s[i]);
    }
    return ans;
}

int gcd (int a, int b) {
    if (a==0) return b;
    if (b==0) return a;
    if (a>b) return gcd(b, a-b);
    return gcd (a, b-a);
}

// -------------- Code --------------
 
int n, m;
vector<vi> d;

int solve(int j) {
    int cnt[n]; for(int i=0; i<n; i++) cnt[i] = 0;
    bool vis[n]; for(int i=0; i<n; i++) vis[i] = false;
    for(int turn = 1; turn <= n; turn++) {
        for(int i=0; i<n; i++) {
            if (vis[i]) continue;
            if(d[i][j] > n+1-turn) {
                vis[i] = true;
                cnt[turn-1]++;
            }
        }
    }
    // for(int i=0; i<n; i++) cout << cnt[i] << " "; enter;
    int permu = 1; int sum = 1;
    for (int i=0; i<n; i++) {
        sum = add(sum, cnt[i]-1, MOD);
        permu = mul(permu, sum, MOD);
    }
    int fact_inv = mod_inv(f[n], MOD);
    // cout << "fact_inv = " << fact_inv; enter;

    int prob_not_touched = mul(permu, mod_inv(f[n], MOD), MOD);
    // cout << "prob not touched = " << prob_not_touched; enter;
    int prob_touched = sub(1, prob_not_touched, MOD);
    return prob_touched;
}

int32_t main() {
    cin >> n >> m;
    d = vector<vi>(n, vi(m, 0));
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            cin >> d[i][j];
        }
    }
    fact(MOD);
    int result = 0;
    for(int j=0; j<m; j++) {
        result = add(result, solve(j), MOD);
    }
    cout << result; enter;
}
