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 = 1000, mod = 998244353;
int n, m, a[N + 1];
int res;
int read() {
  int ret = 0;
  char c = getchar();
  while (!isdigit(c)) {
    c = getchar();
  }
  while (isdigit(c)) {
    ret = ret * 10 + c - '0';
    c = getchar();
  }
  return ret;
}
int calc(int lim) {
  static int f[N + 1][N + 1], s[N + 1][N + 1];
  int ret = 0;
  f[0][0] = 1;
  for (int i = 0; i <= n; ++i) {
    s[i][0] = 1;
  }
  for (int i = 1, j = 0; i <= n; ++i) {
    while (a[j + 1] <= a[i] - lim) {
      ++j;
    }
    for (int k = 1; k <= m; ++k) {
      f[i][k] = s[j][k - 1];
      s[i][k] = (s[i - 1][k] + f[i][k]) % mod;
    }
    ret = (ret + f[i][m]) % mod;
  }
  return ret;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
  }
  sort(a + 1, a + n + 1);
  for (int i = (a[n] - a[1]) / (m - 1), lv = 0; i; --i) {
    int nv = calc(i);
    res = (res + 1ll * i * (nv - lv)) % mod;
    lv = nv;
  }
  printf("%d\n", (res + mod) % mod);
  return 0;
}
