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 sj = 1010;
const int mod = 998244353;
int n, k, a[sj], tot;
long long f[sj][sj], ans, sum[sj][sj], g[100010];
int solve(int x, int y) {
  int l = 1, r = x - 1, tp = a[x] - y;
  if (tp < a[1]) return 0;
  while (l < r) {
    int mid = (l + r + 1) / 2;
    if (tp >= a[mid])
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  tot = (a[n] - a[1]) / (k - 1);
  sum[0][0] = f[0][0] = 1;
  for (int i = 1; i <= tot; ++i) {
    for (int j = 1; j <= n; ++j) {
      int num = solve(j, i);
      f[j][1] = 1, sum[j][1] = j;
      for (int l = 2; l <= k; ++l) {
        if (l > j) break;
        f[j][l] = sum[num][l - 1];
        sum[j][l] = (sum[j - 1][l] + f[j][l]) % mod;
      }
    }
    g[i] = sum[n][k];
  }
  for (int i = 1; i <= tot; ++i)
    ans = (ans + i * (g[i] - g[i + 1] + mod) % mod) % mod;
  printf("%I64d", ans);
  return 0;
}
