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 namespace std;

const int N = 50000 + 9;
const int P = 998244353;

int d[N][25];

long long fact[25], inv_fact[25];

long long inv(long long x) {
    long long r = 1, n = P - 2;
    for (x %= P; n; n >>= 1) {
        if (n & 1) r = r * x % P;
        x = x * x % P;
    }
    return r;
}

long long C(int n, int m) {
    if (n < m) return 0;
    return fact[n] * inv_fact[m] % P * inv_fact[n - m] % P;
}

void solve() {
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            scanf("%d", &d[j][i]);
        }
    }
    long long ans = 0;
    for (int j = 1; j <= m; ++j) {
        int cnt[25];
        memset(cnt, 0, sizeof(cnt));
        for (int i = 1; i <= n; ++i) {
            ++cnt[max(0, n - d[j][i] + 1)];
        }
        long long dp[25];
        memset(dp, 0, sizeof(dp));
        dp[0] = 1; dp[1] = cnt[0];
        for (int i = 2; i <= n; ++i) {
            cnt[i - 1] += cnt[i - 2];
            dp[i] = (dp[i - 1] * (cnt[i - 1] - (i - 1))) % P;
        }
        long long tmp = 0;
        for (int i = 1; i <= n; ++i) {
            int lmt = n - d[j][i] + 1;
            for (int k = 1; k <= lmt; ++k) {
                tmp = (tmp + dp[k - 1] * fact[n - k]) % P;
            }
        }
        ans = (ans + tmp) % P;
    }
    ans = ans * inv(fact[n]) % P;
    printf("%lld\n", ans);
}

int main() {
    fact[0] = inv_fact[0] = 1;
    for (int i = 1; i < 25; ++i) {
        fact[i] = fact[i - 1] * i % P;
        inv_fact[i] = inv(fact[i]);
    }
    int T = 1;
    while (T--) solve();
    return 0;
}