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>
const int N = 1010;
const int M = 2010;
const long long MOD = 998244353;
using namespace std;
int n, k, a[N];
long long f[N][N], sum[N], ans = 0;
void READ() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
}
int GET(int x) {
  for (int i = 1; i <= n; i++) f[i][1] = 1;
  for (int j = 2; j <= k; j++) {
    for (int i = 1; i <= n; i++)
      f[i][j] = 0, sum[i] = (sum[i - 1] + f[i][j - 1]) % MOD;
    int r = 0;
    for (int i = 1; i <= n; i++) {
      while (r < i && a[i] - a[r + 1] >= x) r++;
      f[i][j] += sum[r];
      f[i][j] %= MOD;
    }
  }
  int res = 0;
  for (int i = 1; i <= n; i++) res = (res + f[i][k]) % MOD;
  return res;
}
int main() {
  READ();
  for (int i = (a[n] - a[1]) / (k - 1); i >= 1; i--) {
    ans += GET(i);
    ans %= MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
