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>

#define pb push_back
#define mp make_pair
#define inside sl<=l&&r<=sr
#define outside r<sl||sr<l
#define orta ((l+r)>>1)
#define sol (node+node)
#define sag (node+node+1)
#define inf 1000000009
#define infll 1000000000000000009
#define eps 1e-9
#define ppair(x)  "(" << x.first << ", " << x.second << ") "
#define bas(x) #x << ": " << x << " "
#define prarr(x, n); {cout << #x << ": "; for(ll qsd = 0; qsd < n; qsd++) cout << x[qsd] << " "; cout << "\n";}
#define prarrv(x); {cout << #x << ": "; for(ll qsd: x) cout << qsd << " "; cout << "\n";}
#define ii pair<int, int>
#define endl '\n'
#define mod 998244353
using namespace std;
typedef long long ll;

ll us(ll a, ll b){
    ll ans = 1;
    while (b){
        if (b&1){
            ans *= a;
            ans %= mod;
        }
        a *= a;
        a %= mod;
        b >>= 1;
    }
    return ans;
}

int main(int argc, char* argv[]){
//    freopen("/home/cahid/CLionProjects/untitled/in", "r", stdin);
    cin.tie(nullptr); cout.tie(nullptr); ios_base::sync_with_stdio(false);

    int n, m; cin >> n >> m;
    vector<vector<int> > mat(n);
    for (int i = 0; i < n; i++) {
        mat[i].resize(m);
        for (int j = 0; j < m; j++) cin >> mat[i][j];
    }

    vector<ll> fac(25);
    fac[0] = 1;
    for (ll i = 1; i < 25; i++) {
        fac[i] = i*fac[i-1]%mod;
    }

    ll ans = 0;
    for (int j = 0; j < m; j++) {
        vector<int> arr;
        for (int i = 0; i < n; i++) arr.pb(mat[i][j]);
        sort(arr.begin(), arr.end());
//        prarrv(arr)
        ll ekle = 1;
        for (int i = 0; i < n; i++) {
            if (arr[i] <= i+1) {
                ekle = 0;
                break;
            }
            ekle *= arr[i]-i-1;
            ekle %= mod;
        }
        ekle = fac[n] - ekle + mod;
        ekle %= mod;
        ekle *= us(fac[n], mod-2);
        ekle %= mod;
        ans += ekle;
        ans %= mod;
    }
    cout << ans << endl;

}