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;
long long int table[1001][1001];
int n, k;
int Prev[20000];
long long int arr[20000];
long long int cumm[200000];
void start(long long int gap) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) table[i][j] = -1;
    table[i][0] = 1;
  }
  int ptr = -1;
  for (int i = 0; i < n; i++) {
    while (ptr + 1 < i && arr[i] - arr[ptr + 1] >= gap) {
      ptr++;
    }
    Prev[i] = ptr;
  }
}
void calc(int layer) {
  long long int ans = 0;
  int ptr = -1;
  for (int i = 0; i < n; i++) {
    while (ptr < Prev[i]) {
      ans += table[ptr + 1][layer - 1];
      ans %= 998244353;
      ptr++;
    }
    table[i][layer] = ans;
  }
}
void fill(long long int gap) {
  start(gap);
  for (int i = 1; i < k; i++) {
    calc(i);
  }
  cumm[gap] = 0;
  for (int i = 0; i < n; i++) {
    cumm[gap] += table[i][k - 1];
    cumm[gap] %= 998244353;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  srand(time(NULL));
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &arr[i]);
  }
  sort(arr, arr + n);
  long long int RES = 0;
  for (int i = 0; i < 100001; i++) {
    if (i * (k - 1) <= 100000) {
      fill(i);
    }
  }
  for (int i = 0; i < 100000; i++) {
    RES += (cumm[i] - cumm[i + 1]) * i;
    RES %= 998244353;
    RES += 998244353;
    RES %= 998244353;
  }
  printf("%lld\n", RES);
  return 0;
}
