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 maxn = 1010, mod = 998244353;
long long d[1010][1010], f[101000], ans;
int a[1010], n, k;
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
long long gao(int T) {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= k; j++) d[i][j] = 0;
  int p = 0;
  for (int i = 1; i <= n; i++) {
    d[i - 1][0] = 1;
    while (p < i - 1 && a[p + 1] + T <= a[i]) p++;
    for (int j = 1; j <= min(i, k); j++) {
      add(d[i][j], d[i - 1][j]);
      add(d[i][j], d[p][j - 1]);
    }
  }
  return d[n][k];
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  for (int i = 1; i * (k - 1) <= a[n] - a[1]; i++) f[i] = gao(i);
  for (int i = 1; i * (k - 1) <= a[n] - a[1]; i++) {
    add(f[i], -f[i + 1]);
    add(ans, 1ll * i * f[i] % mod);
  }
  cout << ans;
}
