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;

typedef long long ll;

#define pb       push_back
#define mp       make_pair
#define all(x)   x.begin(), x.end()
#define fi       first
#define se       second
#define cl       clear
#define sep      ' '
#define endll    '\n'

const int maxn  =  1e6   + 10;
const int maxn5 =  5e5   + 10;
const int maxnt =  1.2e6 + 10;
const int maxn3 =  1e3   + 10;
const ll  mod   =  998244353 ;
const ll  inf   =  2e18;

int cnt[maxn5][30];
ll dp[30][30];

ll pw(ll a, ll b){return (b == 0 ? 1LL : (pw(a * a % mod, b / 2) * (b % 2 == 1 ? a : 1LL))%mod);}
int main()
{
    //cout << setprecision(15);
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

    int n, m;
    cin >> n >> m;
    ll fac = 1;
    for(int i = 1; i <= n; i++) fac *= i, fac %= mod;
    for(int i = 0; i < n; i++) for(int j = 0; j < m; j++){
        int a;
        cin >> a;
        cnt[j][n - a + 2]++;
    }
    ll ans = 0;
    for(int i = 0; i < m; i++){
        for(int j = 0; j <= n; j++) for(int k = 0; k <= n; k++) dp[j][k] = 0;
        dp[1][cnt[i][1] - 1] = cnt[i][1];
        for(int j = 1; j < n; j++) for(int k = 0; k <= n; k++){
            if(cnt[i][j + 1] > 0) 
                dp[j + 1][k + cnt[i][j + 1] - 1] += dp[j][k] * (cnt[i][j + 1]); dp[j + 1][k + cnt[i][j + 1] - 1] %= mod;
            if(k > 0) 
                dp[j + 1][k + cnt[i][j + 1] - 1] += dp[j][k] * k; dp[j + 1][k + cnt[i][j + 1] - 1] %= mod;
        }
        ans += (mod + fac - dp[n][0]) % mod; ans %= mod;
    }
    cout << (ans * pw(fac, mod - 2)) % mod << endll;
    return 0;
}