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 ll = long long;

using namespace std;

const ll mod = 998244353;
ll n, m;
ll fact[21];
ll inv[21];

ll solve(vector<ll>& nums)
{
    ll sz = nums.size();
    ll ret = 1;
    for(ll i = 0; i < sz; i++)
    {
        ret = (ret * (nums[i] - i - 1)) % mod;
    }
    ret = ret * fact[n - sz] % mod;
    ret = (fact[n] - ret) % mod;
    if(ret < 0)
        ret += mod;
    return ret;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m;
    vector<vector<ll>> v(n, vector<ll>(m));
    for(ll i = 0; i < n; i++)
    {
        for(ll j = 0; j < m; j++)
        {
            cin >> v[i][j];
        }
    }

    fact[0] = 1;
    for(ll i = 1; i <= 20; i++)
        fact[i] = i * fact[i - 1] % mod;
    
    inv[1] = 1;
    for(ll i = 2; i <= 20; i++)
        inv[i] = mod - (mod / i) * inv[mod % i] % mod;

    ll ret = 0;
    for(ll j = 0; j < m; j++)
    {
        vector<ll> nums;
        for(ll i = 0; i < n; i++)
        {
            if(v[i][j] <= n)
                nums.push_back(v[i][j]);
        }
        sort(nums.begin(), nums.end()); 
        ret = (ret + solve(nums)) % mod;
    }

    for(ll i = 1; i <= n; i++)
        ret = ret * inv[i] % mod;
    cout << ret << endl;

    return 0;
}


