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}
from sys import stdin, stdout
import heapq
from collections import defaultdict
import math
import bisect
import io, os

# for interactive problem
# n = int(stdin.readline())
# print(x, flush=True)

#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def main():
    MOD = 998244353
    #n = int(input())
    n,m = list(map(int, stdin.readline().split()))
    dp_div = [0] * (n+2)
    dp_div[n+1] = 1
    for x in range(n,1,-1):
        dp_div[x] = (dp_div[x+1] * x) % MOD
    arr = []
    for _ in range(n):
        arr.append(list(map(int, stdin.readline().split())))
    ans = 0
    for i in range(m):
        v = [0] * (n + 2)
        for j in range(n):
            x = arr[j][i]
            v[x] += 1
        pos = 0
        cnt = 0
        for j in range(1,n+1):
            if pos == -1:
                break
            while v[j] > 0:
                v[j] -= 1
                if j == 1:
                    pos = -1
                    break
                if cnt == 0:
                    pos = n - j + 1
                    cnt += 1
                    continue
                xx = n - j + 1
                yy = n - cnt
                if xx >= yy:
                    pos = -1
                    break
                pos = ((pos * yy) + (xx * dp_div[yy+1]) - (pos * xx)) % MOD
                cnt+=1
        if pos == -1:
            ans = (ans + dp_div[2]) % MOD
        else:
            while cnt != n:
                pos = (pos * (n - cnt)) % MOD
                cnt += 1
            ans = (ans + pos) % MOD
        y_inv = 1
        for z in range(2, n+1):
            zz = pow(z, MOD-2, MOD)
            y_inv = (y_inv * zz) % MOD
    #print(y_inv)
    print((ans * y_inv) % MOD)













main()
