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;

#define  ll long long
#define ld long double
#define f first
#define s second

ll n , m , mod = 998244353 , a[22][55555]  , dp[22] , P[22][22];

ll fp(ll base, ll exp)
{
    if (exp == 0)
        return 1;
    ll ans = fp(base, exp / 2);
    ans = (ans * ans) % mod;
    if (exp % 2 != 0)
        ans = (ans * (base % mod)) % mod;
    return ans;
}

void calc(vector<ll> v)
{
    sort(v.begin() , v.end() , greater<ll>() );
    memset(dp , 0 , sizeof(dp));
    int j=0 , valid=0;
    dp[0] = 1;
    for(int i=1; i<=n; i++) //dp[i] = probability that first i cities will not have curr point
    {
        while(j < v.size() && n-i+1 < v[j])
        {
            valid++;
            j++;
        }

        if (!valid) return;

        int rem = n-i+1;
        dp[i] = (P[valid][rem] * dp[i-1])%mod;
        valid--;
    }
}

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    for(int i=1; i<=20; i++)
    {
        for(int j=1; j<=20; j++)
        {
            P[i][j] = (1ll*i*fp(j , mod-2))%mod;
        }
    }

    cin >> n >> m;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=m; j++)
            cin >> a[i][j];
    }

    ll ans = 0;
    for(ll i=1; i<=m; i++)
    {
        vector<ll> v;
        for(int j=1; j<=n; j++)
        {
            v.push_back(a[j][i]);
        }

        calc(v);
        ans = (ans + (1 - dp[n] + mod))%mod;
    }

    cout << ans << '\n';

    return 0;
}