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 <atcoder/maxflow.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
 


#include <iostream>
#include <map>
#include <list>
#include <set>
#include <algorithm>
#include <vector>
#include <string>
#include <functional>
#include <queue>
#include <deque>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <iterator>
#include <random>
#include <chrono>
#include <complex>
#include <bitset>
#include <fstream>
 
 
#define forr(i,start,count) for (int i = (start); i < (start)+(count); ++i)
#define set_map_includes(set, elt) (set.find((elt)) != set.end())
#define readint(i) int i; cin >> i
#define readll(i) ll i; cin >> i
#define readdouble(i) double i; cin >> i
#define readstring(s) string s; cin >> s

typedef long long ll;

 
//using namespace __gnu_pbds;
//using namespace atcoder;
using namespace std;

//const ll modd = (1000LL * 1000LL * 1000LL + 7LL);
const ll modd = 998244353;

template<class T>
T fastpower(T x, ll pw, T id = 1) {
    T w = x, res = id;
    while (pw>0) {
        if (pw&1) {
            res = (res * w) % modd;
        }
        pw >>= 1;
        w = (w * w) % modd;
    }
    return res;
}

class Mint {
    public:
      ll val;

      Mint(ll x) {
          if (x<0) {
              x += (1+abs(x)/modd)*modd;
          }
          val = x % modd;
      }

      Mint operator+(const Mint& b) const {     return Mint((val+b.val)%modd);      }
      Mint operator-(const Mint& b) const {     return Mint((val+modd-b.val)%modd);      }
      Mint operator*(const Mint& b) const {     return Mint((val*b.val)%modd);      }
      Mint& operator+=(const Mint& b) {  val += b.val; val%=modd; return *this;  }
      Mint& operator-=(const Mint& b) {  val += modd-b.val; val%=modd; return *this;  }
      Mint& operator*=(const Mint& b) {  val *= (b.val)%modd; val%=modd; return *this;  }
      Mint& operator++()  { ++val; return *this;   }
      Mint operator++(int)  {
        Mint old = *this; // copy old value
        operator++();  // prefix increment
        return old;    // return old value
    }
 
    Mint& operator--()  { --val;   return *this;    }
 
    Mint operator--(int)  {
        Mint old = *this; 
        operator--();
        return old;
    }      

      operator long long() { return val; }
};


class Factorial {
    public:
      vector<ll> bin;

      Factorial(int n) : bin(n+1, 0) {
          bin[0] = 1;
          forr(i,1,n) {
              bin[i] = i*bin[i-1]; bin[i] = bin[i] % modd;
          }
      }

      ll val(int i) {  return bin[i];  }
};

Factorial fact(30);




int main(int argc, char *argv[])   {

    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    uniform_int_distribution<int> rand_gen(0, modd);   // rand_gen(rng) gets the rand no

    ios_base::sync_with_stdio(false);
    cin.tie(0);
 
    cout.precision(12);

//    readint(test_cases);
    int test_cases = 1;
    forr(t, 1, test_cases) {
        readint(n); readint(m);
        vector<vector<int>> d;
        forr(i,0,n) {
            d.push_back(vector<int>());
            forr(j,0,m) {
                readint(aa);
                d.back().push_back(aa);
            }
        }

        Mint ans = m;
        forr(i,0,m) {
           Mint temp = 1;
           vector<int> a;
           forr(j,0,n) { a.push_back(d[j][i]);  } 
           sort(a.begin(), a.end());
           int filled = 0;
           for(auto x : a) {
               if (-filled+(x-1)<=0) { temp = 0; }
               temp *= -filled+(x-1);
               ++filled;
           }
           temp *= fastpower<ll>(fact.val(n) ,modd-2);
           ans -= temp;
        }
        cout << ans << endl; 

    }

    return 0;
}

