Let's call beauty of an array b_1, b_2, …, b_n (n > 1) — min_{1 ≤ i < j ≤ n} |b_i - b_j|.

You're given an array a_1, a_2, … a_n and a number k. Calculate the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.

A sequence a is a subsequence of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements.

Input

The first line contains integers n, k (2 ≤ k ≤ n ≤ 1000).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^5).

Output

Output one integer — the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.

Examples

Input


4 3
1 7 3 5


Output


8

Input


5 5
1 10 100 1000 10000


Output


9

Note

In the first example, there are 4 subsequences of length 3 — [1, 7, 3], [1, 3, 5], [7, 3, 5], [1, 7, 5], each of which has beauty 2, so answer is 8.

In the second example, there is only one subsequence of length 5 — the whole array, which has the beauty equal to |10-1| = 9.
#include <bits/stdc++.h>
using namespace std;
const int N = 1010, mod = 998244353;
int n, k, d[N], f[N][N], s[N][N];
inline int read() {
  int x = 0, p = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * p;
}
void init() {
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) {
    d[i] = read();
  }
}
int get(int x) {
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= n; ++j) {
      f[i][j] = 0;
      s[i][j] = 0;
    }
  }
  for (int j = 1; j <= n; ++j) f[1][j] = 1;
  for (int i = 1; i <= k; ++i) {
    int l = 1;
    for (int j = 1; j <= n; ++j) {
      while (d[j] - d[l] >= x) ++l;
      f[i][j] = (f[i][j] + s[i - 1][l - 1]) % mod;
      s[i][j] = (s[i][j - 1] + f[i][j]) % mod;
    }
  }
  int ret = 0;
  for (int i = 1; i <= n; ++i) {
    ret = (ret + f[k][i]) % mod;
  }
  return ret;
}
void solve() {
  sort(d + 1, d + n + 1);
  int ans = 0, lim = d[n] / (k - 1);
  for (int i = 1; i <= lim; ++i) {
    ans = (ans + get(i)) % mod;
  }
  cout << ans << '\n';
}
int main() {
  init();
  solve();
  return 0;
}
