E. Assimilation IVtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputMonocarp 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!InputThe first line contains two integers n and m (1 \le n \le 20; 1 \le m \le 5 \cdot 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 \le d_{i, j} \le n + 1) is the distance between the i-th city and the j-th point.OutputIt 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 \frac{x}{y}. Print this fraction modulo 998\,244\,353, i. e. value x \cdot y^{-1} \bmod 998244353 where y^{-1} is such number that y \cdot y^{-1} \bmod 998244353 = 1.ExampleInput
3 5
1 4 4 3 4
1 4 1 4 2
1 4 4 4 3
Output
166374062
NoteLet'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 \frac{4 + 3 + 3 + 3 + 3 + 3}{6} = \frac{19}{6} or 19 \cdot 6^{-1} \equiv 19 \cdot 166374059 \equiv 166374062 \pmod{998244353}
3 5
1 4 4 3 4
1 4 1 4 2
1 4 4 4 3

166374062 
['combinatorics', 'dp', 'math', 'probabilities', 'two pointers', '*2100']