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}
import sys,io,os
try:Z=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
except:Z=lambda:sys.stdin.readline().encode()
Y=lambda:map(int,Z().split())
M=998244353
try:
    import __pypy__
    int_add=__pypy__.intop.int_add
    int_sub=__pypy__.intop.int_sub
    int_mul=__pypy__.intop.int_mul
    def make_mod_mul(mod=M):
        fmod_inv=1.0/mod
        def mod_mul(a,b,c=0):
            res=int_sub(
                int_add(int_mul(a,b),c),
                int_mul(mod,int(fmod_inv*a*b+fmod_inv*c)),
            )
            if res>=mod:return res-mod
            elif res<0:return res+mod
            else:return res
        return mod_mul
    mod_mul=make_mod_mul()
except:
    def mod_mul(a,b):return(a*b)%M
def mod_add(a,b):
    v=a+b
    if v>=M:v-=M
    if v<0:v+=M
    return v
def mod_sum(a):
    v=0
    for i in a:v=mod_add(v,i)
    return v
n,m=Y();d=[[*Y()]for i in range(n)];s=0;inv=1
for i in range(n):inv=mod_mul(inv,i+1)
inv=pow(inv,M-2,M)
def f(l):
    v=1;n=len(l);p=n-1;c=0
    for i in range(n):
        while p>=0 and l[p]>n-i:c+=1;p-=1
        if not c:return 0
        v=mod_mul(v,c);c-=1
    return v
for i in range(m):
    l=sorted(d[j][i]for j in range(n))
    p=mod_mul(f(l),inv);s=mod_add(s,1-p)
print(s)